non-abelian groups whose groups of isomorphisms are abelian

Annals of Mathematics

Non-Abelian Groups Whose Groups of Isomorphisms are AbelianAuthor(s): C. HopkinsSource: Annals of Mathematics, Second Series, Vol. 29, No. 1/4 (1927 - 1928), pp. 508-520Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1968020 .

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NON-ABELIAN GROUPS WHOSE GROUPS OF ISOMORPHISMS ARE ABELIAN.*

BY C. HOPKINS.

Introduction. In the appendix of Hilton's Finite Groups (1908), page 233, the question whether a non-abelian group can have an abelian group of isomorphisms occurs among "a few interesting questions still awaiting solution". A non-abelian group of order 64 whose group of isomorphisms is abelian and of order 128 was later constructed by G. A. Miller.t No other discussian of the problem appears in the literature of Mathematics. In fact, an exhaustive analysis of the properties of a non-abelian group whose group of isomorphisms is abelian seems to involve some difficulty, for the reason that little is known concerning isomorphisms of a non-abelian group.

The nature of an abelian group whose group of isomorphisms is abelian was determined by G. A. Miller,+ who proved that a necessary and sufficient condition that an operation of the group of isomorphisms of an abelian group be invariant under this group is that it should transform every operation of this abelian group into the same power of itself. From this it is obvious that the only abelian groups whose groups of isomorphisms are abelian are the cyclic groups. In what follows we shall develop a few necessary conditions which must be satisfied by a non-abelian group whose group of isomorphisms is abelian.

1. The fundamental theorem. Wle assume a non-abelian group G which is restricted solely by the hypothesis that its group of isomorphisms I is abelian. Since the central quotient group of G is simply isomorphic with the group of inner isomorphisms of G, it is clear that every commutator of G is invariant under G. The group G is accordingly the direct product of its Sylow subgroups.? Since every Sylow subgroup of G must correspond to itself in any isomorphism of G with itself, we may confine our- selves to the case where G is of order pm.

For convenience we shall introduce the following notation: the symbol G shall consistently denote a non-abelian group whose group of isomorphisms I

* Received January 22, 1928. t G. A. Miller: Mess. of Math., vol. 43 (1913-1914), p. 124. t G. A. Miller: Trans. Amer. Math. Soc., vol. 1 (1900), p. 397; vol. 2 (1901), p. 260. ? Buruside: Theory of Groups of Finite Order, 1897, p. 115.

508



GROUPS OF ISOMORPHISMS. 509

is abelian; operations in G and operations of I shall be denoted by lower- case and capital letters respectively; an isomorphism of G with itself will often be called an automorphism, for the purpose of distinguishing it from an isomorphism of G with a subgroup of G. The operations s-' t-' st shall be called a commutator of G when both of its elements are in G; the operations s-1 T-1 s T shall be referred to as a multiplier of G when one of its elements T is in I. It will be noted that these forms are not mutually exclusive.

Let T1 and T1' be two distinct operations of I and let s be any operation of G. Suppose that

T71 sTi = sri and T1 sT2 S S2.

Also

'i 1r2 T1 = T2 T12 and T71 rl T2 = iY21.

Now Z1 7T-1 sT2 T T= 1 S T2 T =: Sfl Y2 T12,

and T27' T7 'sT T2 = T7'sr, T2 - ST2TlT2,.

Since I is abelian, TT2 = 2 TT,

Therefore TiT2 Ti12 = 2 T2 Y2i

If ri is the identity, then the relation above reduces to

T2Ti2 = T2,

which implies that r12 is the identity. From this we have the theorem that every operation of I which is commutative with an operation s of a is commutative with every conjugate of s under I.

2. Commutators and multipliers. We shall now prove that every commutator of G is a characteristic operation: i. e. an invariant operation under the holomorph of G. Suppose that s and t are any two non- commutative operations of G and let s-1 ts = tc. We shall assume that c is not characteristic: i. e., we shall assume an outer isomorphism T which effects the following correspondences:

s--s-90 Y to- thou r, c cco

Since conjugates must correspond,

-s-1 ts -s-1 s-1 t t ss



510 C. HOPKINS.

Since products must correspond, t c t to c co.

We assumed that s-1 ts = tc; therefore

s0 s1 t0 ss= It0cc0.

NNow there is an inner isomorphism S in I which transforms the operations of G according to s; in symbols

S-'tS= tc. Now

S-1 T-1 tTS =S-1 ttoS to tos,

where t = to 1S'1 t S, and

T-1S- ST --- T-1 tcT tto co. Since I is an abelian,

ST TS. Therefore

to CCo Cto tos,

and from the fact that c is in the central of G we obtain the relation tos = co; that is

S-1 to S = toco

But S and s transform the operations of G in the same way: therefore

a) s-tos =- toco.

Since s1 tIs = tc it follows that t'1st = sc'. Now there exists an inner isomorphism which transforms the operations

of G according to t. By an argument parallel to that employed above we can show that

b) t-'so t= soco1.

Relations a) and b) will be used to simplify the equation

C) S-l S-1 t to SSO = t to c cO. Now

so-S-1 ttoss O so-1 ses-1 t oS stss0 s - so ~ tctc C so.

Substituting in c), SO-t Ctoo so = = to cco.




Both c and c0 are invariant under G; therefore

so 1 t to so t too sj't0 s0=t0

From the theorem of ? 1 we know that so must be commutative with t, since it is commutative with its conjugate tto under T. Since t-1 sot

s cr1, from relation b), it follows that co must be the identity. In other words, every commutator of G is characteristic.

From the fact that c is characteristic, relations a) and b) reduce to s-1 to s = to and t-1 so t = so respectively. If s and t are commutative, then, by the theorem of ? 1, s must be commutative with every multiplier of itself and of t. The obvious conclusion is stated in the corollary: every multiplier of G is invariant tunder G.*

3. Non-divisibility of G. A group which is the direct product of two groups is said to be a divisible group. We shall now show that G cannot be divisible with one of its factors abelian. Let G be the direct product of an abelian group A and a non-abelian group N. It is clear that A must be cyclic; for a non-cyclic abelian group always admits isomorphisms which are not invariant under its group of isomorphisms. We assume, then, that A is generated by t of order pa. Let s, of order pfn, be an operation of highest order in N.

If a < n, we can construct an automorphism T, of G which will transform s into st; we can also find a second automorphism T2 for which

a) T7 'sT2 s and T7 tT2 = ts', where s' = tP, if a>1, or, if t is of order p, where s' is an invariant operation of order p from N.

If a> n, we can construct an automorphism T1' which will transform s into s tP 8; we can also establish an automorphism T2' for which

b) T2' sT2' = s and T2' tTo T t s', where s' = tP, if n>1. If n= 1, then, from the assumption that s is an operation of highest order in N, it follows that every operation in N is also of order p. Since a group in which every operation is of order 2 must be abelian, p must be an odd prime for the reason that NI is non-abelian. If p 2, then we may select t as the multiplier s' of t tinder Tot.

In every case, then, we can find an automorphism (T2 or TY') which is commutative with s but not with all the conjugates of s under the group of isomorphisms of G. From this contradiction of the result expressed

* In view of the fact that the group generated by the commutators of G is a subgroup of the group generated by the multipliers of G, it is not surprising that the multipliers are subject to a weaker restriction. In the example given by G. A. Miller (loc. cit.) there is a multiplier of order 4 which is not characteristic.



512 C. HOPKINS.

in the theorem of Art. 1 we have proved that the group G cannot be a divisible group with an abelian group as one of its factors.

4. Generating operations of G. It is a well known fact that the number of independent generators of a group of order pm is an invariant of the group.* In the case of an abelian group of orderpm its independent generators may be so selected that the group is the direct product of the cyclic subgroups generated by each of its independent generators. A direct analogy does not exist in the case of the group G which we are consi- dering. However, we can show that, if p is an odd prime, then any set of independent generators (s1, 52 .* 5* ) SO of the group G can be replaced by a set of independent generators (si', s', *, s,) which have the property that no two of them shall have any powers in common except the identity.

Let us suppose that the orders of s1, s2, *, s. are peg, pal . an

respectively, where a1 ? M, > ... > an. We shall assume that p9' is the lowest power of p for which S,2 SP# 'IP s... are each in {is} while the p#2-1 power of at least one of the generators, say si, is not in (511. Since ai < a, there must be a power of s , say s a whose pf2 power is the inverse of sO'. Then the order of s? sal is pi,. For

(sa pAo )a, SiP I, s lSa s l5s2 * Sp S.l St. P*p

If Si Sa, S-i = ,C 1f s1s s -sc, then

-S~l~P =as csa, c' ... 5a, cp~SpA (Si Sil 1 1 1

Since the commutators of G are invariant

aS p1') a

- I PP2SA 2 pP92 (pAl 1 (a1)PPI = 5l~~~~~ sPPA c2

Now the order of c cannot exceed p I. f p> 2, then +1) = 1. But (s1l)jpP -P2. Therefore

(si sa'),' - S. _pP2 = 1.

If we denote the product s saI by s2, we see that the generator si may be replaced by the operation s' of order p9, which has no powers in common with sj, except the identity. This last statement is obvious from the fact that one of the constituents of s' is in {Is} while the other is not.

Suppose now that p9' is the lowest power of p for which the pf power of each of the (n -2) generators obtained by excluding si and s' is in

* G. A. Miller: Trans. Amer. Math. Soc., vol. 16 (1915), p. 21.




either {si} or {s,}, while the p3-` power of at least one of them, say si, is neither in {si} nor in {s'}. We may note that if s8Pl? is in both {si} and {s(} then 8P"a = 1; otherwise, (s8l and {s'} would contain operators in common other that the identity. If sP" is in {si} and is not the identity, J then the argument employed above may be used without modification to obtain an operation s' of order pl which has no powers in common with s1 except the identity; and it is further clear that the operation s which is of the form s, sl can have no powers in common with s' except the identity. If se;' is in {s'} and is not the identity, we might reasonably ask whether it would always be possible to find a power of sA whose pig power would be the inverse of S73. It is obviously possible to do so if the order of sj does not exceed the order of s'. And if aj > f2, then, remembering that sPp' is in {s'} and is not the identity, it follows that sP$ is necessarily J 2I the first power of sj which is in Is, }, a result which contradicts our assumption that the p :2 power of every generator is in {sj}. Regardless of whether se' is in {sjj or in {s>t, we can always find an operation s' of order pig which has no powers in common with either s1 or sA except the identity; and this operation s3 can be used as a generator in place of sj.

By a repetition of this argument it can be proved that if p> 2, the generating operations of G may be selected in such a manner that they will have no powers in common except the identity.

Let us see what statement can be made regarding the selection of the generators of G when p 2. We remember that in the demonstration

above it was necessary to show that c2 = 1, where c is a power of the commutator of sj and s,. The proof depended on the fact that if p is an odd prime, then pie2 + 1 is divisible by 2. When p = 2 this is no

longer the case; consequently it is possible that c2 is not the identity.

Let (s3, S2, ., S.) be a set of independent generating operations of G of orders 2a12as ..* 2an respectively, where a, > a2 >?. a,,. Let 2# be

the lowest exponent for which S22 , s' * * are all in {s1}, while the 2#-1 power of one of the set ($9, . a, s,) say sl"- , is not in {sj. Then there must be some power of sj, say s{2a which is the inverse of s229. (Since the following argument does not depend upon the value of k, we shall take k equal to unity.) It is evident that a, - a2 = a -,f.

We shall now determine the order of s2 8~a -4, assuming that s2 s1 s2 m = sl cl where the order of c obviously cannot exceed 2fl. By raising s2 sIG" to its m-th power, we obtain the following relations:

33 Vol.29



514 C. HOPKINS.

(S2 2CXP')m -_ s2aP S-1 . 2aP S-2.5m s2a-P M 21 S2 S1 S2 * 2 1i 2 S2 I S1 S2

- Sm2aJ Sm . c2P ma(mi) (m+1)/2 1 2

If we set m equal to 2f we see that (S2 S2-a) reduces to c2c' (2+1), since -~la 812P. If ce> , then C2 (2+1)

- 1 and the order of s2sia is 00.

hence we could replace s2 in (s *,*, s) by s1 - =2 Q where s1 and s

obviously have only the identity in common. If ac == and the order of c is less than 2fl, the conclusion of the preceding sentence still holds.

If, now, a = f and the order of c is equal to 2#, it is clear that the order of S2 S4aP = SS equals 2#12, since s2 is the inverse of sc. Even here we can take S2 SI for the operation s2 which is to replace S2, unless Si and s2 s8 have in common an operation of order 2, namely C21''(291). If s, and s2 s1 have in common this operation of order 2 and if a + 1 < a, where pa is the common order of s, and s2, then we can choose for s2 the operation s2Si+2a/3l of order 21 which has no powers in common with sl except the identity. That the order of S2 S1+2a4# is 2# may be seen from the following equations:

1+2 a--1)2P - S~2"+2 -1 93 5c(1+2apl) 2P (2P+l) 2~ . 1 2

2a-1 2fr-1

(We remember that a -,8 > 1; hence c2a-2 1. Also, s2 and c2P' are the same operation of order 2.)

To recapitulate: Let (Si, S2, *--, SO) be a set of independent generating operations of G of orders 2a1 2a2, *- 2a respectively, where al ? a2 ?*..?an;

if 28 be the lowest exponent for which s'2P, **, s. are each in {sT} while the 2#-' power of one of the set (s2' **, S,) say S2.4 is not in {S1}; then, either (1) it is possible to find an operation s' of order 21 which has no operations in common with sI except the identity and which can replace sj in the given set of generating operations, or (2) s, and sj are of the same order 2a, their commutator c = 571 Si S,

l is of order 2a-1, the operations s1, sj have in common only the subgroup generated by the commutator C2a-2 of order 2.

From this point on the discussion follows the lines developed at the corre- sponding stage in that case p > 2, with the necessary modifications suggested by the results in the preceding paragraph. Finally we arrive at the following theorem, which logically falls into two sections according to the value of p.




If p >2, a set of independent generating operations of G can be found which shall possess the property that no two of them shall have in common any powers except the identity.

If p = 2, a set of independent generating operations of G can be found which shall possess the property that any two of them have in common at must a commutator of G of order 2.

A set of generating operations for which this property holds shall be referred to as a restricted set of generating operations. In what follows it will be assumed that when reference is made to a set of independent generating operations, it is to a set selected in this manner.

5. The order of I. Since the group of inner isomorphisms of G is simply isomorphic with the central quotient group of G, it is clear that its order is a power of p. In this section we shall show that the order of every isomorphism of G is likewise a power of p. We shall employ a method of attack which is suggested by the theorem* that every automorphism of an abelian group A may be obtained (1) by making A isomorphic with one of its subgroups or with itself in such a manner that no operation besides the identity corresponds to its inverse, and (2) by making each operation of A correspond to itself multiplied by the operation which corresponds to it in this isomorphism.

Since every multiplier of G is invariant in G (cf. ? 2), with every automorphism of G there is associated an isomorphism of G with a subgroup K of the central of G. Let s represent any operation of a set of independent generators of G and assume that s corresponds to sc under the automorphism T. If we denote by H the subgroup of G which is composed of those operations which are commutative with T and by K the subgroup of the central of G with which G is isomorphic under the automorphism T, then the correspondence of the operations of G under T may be exhibited by the following scheme:

A) H -Hy 1,

it2 Xt.-I ct2

HsA Hs c

where 1, c2, * ., c are operations of K. If T is commutative with c, it is evident that the exponent of the first

power of T which is commutative with s is a power of p. Let us suppose, then, that T and c are not commutative; that is, we assume the relations

* Miller, Blichfeldt, and Dickson: Finite Groups, p. 103.



516 C. HOPKINS.

T-1sT - sc, T-1cT = cy (r t 1).

By hypothesis, s is a generating operation of G; consequently we may take c as a generating operation of the abelian subgroup K. From the automorphism T we may now construct an automorphism T' by replacing (vide scheme A) in K the multiplier c by the identity, which will have the effect of enlarging H by the adjunction of s and of other operations which under T are multiplied by powers of c. Since T' is commutative with s, it must not be commutative with c, by the theorem of Art. 1. This demands that r be a power of c. Then, since r = c#, it follows that the cosets Hc and Hs# associated with the automorphism T must be identical (vide scheme B). This implies that c s= 0, where r is an operation of H.

B) H SH. 1,

Hc = sHccP Hc.r,

Hs - Hs c.

If y and c are of the same order, then c and s must likewise have a common order; if, in addition, s is a non-invariant operation of G, then c, which equals sO r, must also be a non-invariant (since s# does not occur in H). This, however, contradicts the theorem that the multipliers of G are invariant. If s is an invariant in G and c is of the same order as s, then, since the operations of the commutator subgroup of G are characteristic, the first power of s which is in the commutator subgroup must be the identity; the group G would then be divisible with {s} for one of its factors, which is contrary to the result established in ? 3. We see, therefore, that y is of lower order than c; as a consequence, from the relation c sfl r, it follows that c has as a constituent a power of s of lower order than the order of s.

We may note that the operation r of H which occurs in the equation c = s# r must be a characteristic operation of G. Suppose that there exists an automorphism T1 of G in which r corresponds to rr', (r't 1). It may be assumed that s and r have no powers in common except the identity; consequently, from the automorphism T1 we can construct a second automorphism T2 in which r corresponds to rr' and s corresponds to itself. But T2 is commutative with s and not with c. Hence r must be a characteristic operation of G.

To recapitulate: If T- s T = sc and T-1 c T = c, then the exponent of the first power of T which is commutative with s is a power of p; if T-1sT sc and T-1cT cy, then c is of lower order than s andr




is of lower order than c. Combining these two results, we may assert that the exponent of the first powcer of any automorphism of G which is commutative with a generating operation of G is a powter of p.

It is now easy to show that the order of every automorphism of G is a power of p. Since the commutators of G are invariant, any operation t of G may be written in the form

xi12 5n0 t sl s2 ... son s0

where ss, s2, s*,S are a set of generating operations and so is in the commutator subgroup of G. (It is understood that sl, s2, *.., 5sn constitute a restricted set of independent generators of it [? 4]). Let T be any automorphism of G and let TP ', TP2, *.., TPn be the first powers of T which are commutative with sxl, s2 *2 * s5X respectively. It is easily seen 1 2 n~~~a that t and TP are commutative, where p is the largest of the exponents pI, pa2 * a*, pa. Therefore the exponent of the first power of T which is commutative with t is a divisor of pai and, consequently, the order of T is a power of p. In other words, the order of the group of isomorphismns of G is a power of p.

From the discussion above it is obvious that the order of any automorphism of G it at most equal to the order of an operation of highest order in the subgroup K associated with T, (vide scheme B). That is, the order of every automorphism of G is a divisor of the order of an operation of highest order in the central of G.

6. Restrictions on the multipliers. Let s be any one of a set of restricted generating operations of G. In ? 5 it was proved that if T-'sT= sc and T-cT= cy, where yr 1, then r = ck2P2 and c =sk"r,

where r is a characteristic operation of G. In the proof of the relation c = skPpr no use was made of the assumption r + 1. Therefore we have in every case the result: Every multiplier of a generating operation * s of G is the product of a characteristic operation of G and a power of s which is invariant under G.

In ? 5 it was also proved that the order of the group of isomorphisms I of G is a power of p; consequently, the number of conjugates of s under I is a power of p, say pk. Since every operation in I which is commutative with s is commutative with every conjugate of s under I, it follows that 1, regarded as a substitution group on the pm letters of G, must contain a subgroup J of order pA having a transitive constituent of degree pi under which s is transformed into all of its conjugates under I. Let us denote

* Vide p. 515.



518 C. HOPKINS.

the operations of J by 1, T2 *, T A and let us assume the following relations:

TT's T =s c, TJ1's Tj sc, TF1 cj T=-cjcij, T7 ci Tj cs .

By the argument which was used in ? 1 it can be proved that cocjcij = cjc,'cji. Since the multipliers of G are in the central of G (cf. ? 2) it follows

that c = cji. Suppose that TC7 s Ta = sca, where ca is a characteristic operation of G; then Caj= ci = 1, which implies that Ta is commutative with every multiplier of s under I.

If T,,-,s'Ta = s Ca and Ti' s Tp = scp while cas + 1, then cp must contain as a constituent a power of s, since every multiplier of s is of the form skPr where r is a characteristic operation of G. Let Cp = skpPr. Now

Ta' T-' s T T a i1sskPr Ta scas -Pi r while

T' T-' sTaT = T-'scaTp sskPI rcacpa. Since Ta T = Tp Ta,

Ca 5k -P US kP4rCaCa

whence Ca - c'kP. From the relation cpa = cap and from the fact that cp is not characteristic [Cap f 1, by assumption] one sees that Ca must involve a non-characteristic power of s; i. e.,

Ca = skP' r'

where r' is a characteristic operation. It is obvious that J P-ntains an operation Tr which transforms s into sr. Then

Tr1 Ca Tr = Tr-' 5skP r' . Tr skP r' . rk P

In the preceding paragraph it was proved that an automorphism which transforms s into itself multiplied by a characteristic operation of G must be commutative with every multiplier of s under I. Therefore rkp = 1. Similarly it can be proved that rkP' - 1, r'k'p= 1, and r'kPP - 1. From these relations and the relations Ca = Poa r', cp-skP' r, Cpa ck Cap = Cpa, we see that Cap and cpa have the common value sk/'P",. This result may be stated in the following form: if the multipliers of a generating operation s are not all characteristic, then I must contain an operation Ta which transforms s into s * skp r and transforms sky into itself multiplied by skP , where k is prime to r, a >O, sk6J is in the central of G, s7 t 1, and r is a characteristic operation of G.



GROUPS OF ISONIORPHISMS. 519

It is clear that the multipliers 1, c2, *.., cPA associated with the operations 1, T2, , TA 2of J are all distinct. One readily sees that, if s corresponds to scex under Ta and to sc: under T#, then J must contain an automorphism under which s corresponds to scac:. WVe note that this automorphism is the product Ta To if, and only if, c,, and T: are commutative. Since the product of two multipliers is itself a multiplier, we conclude that the multipliers of s form a subgroup r of order pl. Evidently the subgroup r contains the multipliers cij. It is clear that F and J are simply isomorphic if the multipliers of s are characteristic. Suppose that F contains a multiplier which is not characteristic. By the theorem of the preceding paragraph we know that J must contain an operation Ta for which

T 1 s Ta = ssCPa r and Tj 1 slcPa r Ta = SPa A_ Sk2P2a

where so 2P" is not the identity. It has been proved above that the order of every characteristic multiplier is a divisor of pa if sa is a multiplier of s which is not characteristic. We see, then, that the order of Ta is a divisor of the order of skPa Suppose that Ta is of order* pb. Thent

T-PB s TPB~ S(l+Pcl)P,

Now

(1p~b =b+ba1 1-1 a+..+(pb 1)! p(r-l)a+1 +~~ ~ +p2 p + * *~ I+ + (P )!P + ]

If the order of Ta is less than the order of sP8, then the quantity Q in the brackets must be divisible by p. If p is an odd prime, then the coefficients of the powers of p contained in Q are all integers and Q is of the form 1+kp. If p=2anda>1, thenQ 1mod.2. If a =1, then (14+ 2)2" is represented by the following expansion:

-1 (1 + 2)2b 1 + 2b+1 [1 + (2b - 1) + 2b-I(2b-1 -1) 22

+ b -

12 )(93 + 61(2b-1 -1) (2b -3) 2

2-1 + 2 I (2b 1) (2b - 3) (2b-2 - 1) 22 + terms divisible by 22].

* Since I is abelian and s is a generating operation of Q, it may be assumed that Ta is commutative with the remaining generating operations of the set to which s belongs. Then the order of Ta is the first power of Ta which is commutative with s.

t We note that rP7 = 1. Since the following discussion does not depend on the value of k, we shall, for convenience, take k = 1.



520 C. HOPKINS.

We see that in this case Q is even. If the order of s is greater than 4, then pb, the order of Ta,, divides one half the order of S2 =- se and, of course, one-fourth the order of s. We see, therefore, that only in the exceptional case where p = 2 and a = 1 can the order of Ta be less than the order of the multiplier of To and s. The general result may be stated in the following theorem: if p is an odd prime, then the subgroup J of the group of isomorphisms I of G which is composed of those operations of I which transform a generating operation s of 0 into all its conjugates and are commutative with the remaining generating operations of G is simply isomorphic with the subgroup of the central of G which is generated by the multipliers of s. The same statement can be made when p = 2 if either of the following conditions is satisfied: (1) the order of s does not exceed 4; (2) there is no operation in J which transforms s into s . s2k where k = 1 mod 2. If there is an operation T in J which transforms s into so ok and if the order of s is greater than 4, then the order of T divides one-fourth of the order of s.



non-abelian groups whose groups of isomorphisms are abelian

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