MATH 111A: GROUP THEORYASSIGNMENT 4: SOLUTIONS

Exercise 1 Let G be a group and let H be a subgroup of G. Let g G and let h Hg. Provethat Hg = Hh.Solution 1 Since h Hg we can write

h = h1g h11 h = gfor some h1 H. Let h2h Hh, with h2 H. Then h2h = (h2h1)g Hg, and weve thus shownthat Hh Hg. On the other hand, if h3g Hg for h3 H, then h3g = (h3h11 )h Hh and weget also Hg Hh. Thus Hg = Hh and were done.

Exercise 2 Find an example of a group G, a subgroup H G, and two elements g, h G suchthat Hg = Hh but such that gH 6= hH.Solution 2 We must take a nonnormal subgroup. So lets start with the smallest nonabelian

group G = S3, and see if any nonnormal subgroup will work: for example, lets try H = {e, (12)}.This is not normal because (123)(12)(123)1 = (123)(12)(132) = (23) 6 H. We have

H(123) = H(23) = {(123), (23)},while

(123)H = {(123), (13)},(23)H = {(23), (132)} 6= (123)H.

Exercise 3 Let

G =

{(a b0 c

) GL2(R) | ac 6= 0

},

H =

{(1 x0 1

) G

}.

Note that H is a normal subgroup of G.

(a) Prove that two cosets (a b0 c

)H,

(x y0 z

)H

are equal if and only if a = x and c = z.(b) Use the observation in part (a) to define an isomorphism

G/H = R R.Solution 3

(a) Note that (a b0 c

)H =

{(a b+ ax0 c

) GL2(R)

}.

1

2 ASSIGNMENT 4

Every matrix in this coset has a and c on the diagonal. Thus, the only way two such cosetscan be equal is if the diagonal entries are all equal. This proves that if(

a b0 c

)H =

(x y0 z

)H,

then a = x and c = z. On the other hand, if a = x and c = z then(x y0 z

)1(a b0 c

)=

1

ac

(c y0 a

)(a b0 c

)=

1

ac

(ac c(b y)0 ac

)=

(1 (b y)/a0 1

) H.

Hence these two matrices define the same left coset of H. This concludes the only if partof the proof.

(b) By part (a) the map : G/H R R defined by

((a b0 c

)H

)= (a, c)

is a bijection between the left cosets of H mod G and the elements of the cartesian productR R. We only have to check that is a group homomorphism. By definition theidentity of G/H goes to (1, 1), which is the identity of R R. To see that respectsmultiplication we check:

((a b0 c

)H

(x y0 z

)H

)=

((ax ay + bz0 cz

)H

)= (ax, cz)

= (a, c)(x, z)

=

((a b0 c

)H

)

((x y0 z

)H

).

This concludes the proof.

Exercise 4 Let

D8 = {1, , 2, , 7, , , 2, , 7}denote the dihedral group of order 16, where 8 = 2 = 1 and = 7. Let H = {1, 2, 4, 6}and note that H is a normal subgroup of G.

(a) Show that D8 contains exactly 9 elements of order 2.(b) Write down the subgroup lattice for D8.(c) Show that there is a partition

D8 = H H H H.(d) Prove that D8/H is isomorphic with (Z/2Z) (Z/2Z).(e) Use this to write down the subgroup lattice of D8/H. Do you recognize this as a sublattice

of the one found in part (b)?

Solution 4

ASSIGNMENT 4 3

(a) We have seen that all elements of the form n are of order 2, and there are 8 of these. Wealso note that ord(d) = 8/ gcd(d, 8). The only way this can be 2 is if d = 4, hence 4 givesthe ninth and final element of order 2 in D8.

(b) The cyclic subgroups of correspond to the divisors 1, 2, 4 and 8 of 8. These are thegroups

8, 4, 2, .All other cyclic subgroups are the ones of order 2 generated by the reflections. To findmore subgroups we can consider pairing the cyclic subgroups. Pairing two powers n andn gives another subgroup of , and so weve accounted for it already.

What if we pair two reflections, say n and m with m 6= n? If H is a subgroupcontaining them then it also contains

nm = mn.

If m n 1, 3, 5, 7 (mod 8) then mn generates . In this case the group containingn and m must be all of D8, since D8 is generated by and any reflection, not just .If m n 2, 6 (mod 8) then mn generates 2. In this case we find that n and mgenerate one of two possible copies of D4 inside D8: (assume wlog n < m)

n, m ={{e, 2, 4, 6, , 2, 4, 6} if (n,m) = (0, 2), (0, 6), (2, 4), (4, 6),{e, 2, 4, 6, , 3, 5, 7} if (n,m) = (1, 3), (1, 7), (3, 5), (5, 7).

The last possibility is that m n 4 (mod 8) and then mn generates 4. Similar toabove, in this case we find 4 copies of D2 = (Z/2Z)2 inside D8:

n, m =

{e, 4, , 4} if (n,m) = (0, 4),{e, 4, , 5} if (n,m) = (1, 5),{e, 4, 2, 6} if (n,m) = (2, 6),{e, 4, 3, 7} if (n,m) = (3, 7).

At this point weve found all the subgroups, but were not done our argument! We stillhave to show that no other subgroups are possible. Lets continue by seeing what happenswhen we pair a rotation n with a reflection m. In this case one gets one of the dihedralgroups encountered above, so this case contributes nothing new. By Lagrange, if we addanything new to a copy of D4 in D8 then it must be the whole group, since the order ofevery subgroup has to divide 16, and D4 has order 8 so there are no nontrivial subgroupsabove any copy of D4 inside D8. It remains to consider what happens if we add a newelement to one of the four copies of D2 that we found inside D8. If we add one of theelements 2 or 6 to a copy of D2, then one obtains also

2 and obtains one of the 2 copiesof D4 encountereed already. Adding ,

3, 5 or 7 to one of the D2s generates the wholegroup. Now we only have to consider what happens if we add a reflection to one of thecopies of D2. Depending on what reflection you add, it is not hard to see that youll get aD4 or the entire D8. For example, consider

H = {e, 4, , 4}.If we add 2 or 6 to this group then we obtain the first copy of D4 seen above. If weadd , 3, 5 or 7, then the resulting subgroup will contain all of and hence mustequal D8. This treats all of the relfections not already contained in H.

4 ASSIGNMENT 4

Thus, weve shown that all subgroups generated by 3 elements are contained in the listweve found already, so were finished! The subgroup lattice looks like:

D8

jjjjjjjj

jjjjjjjj

jjjj

SSSSSSSS

SSSSSSSS

SS

2,

jjjjjjjj

jjjjjjjj

jjj

TTTTTTTT

TTTTTTTT

TTT 2,

TTTTTTTT

TTTTTTTT

TT

kkkkkkkk

kkkkkkkk

k

4,

HHHH

HHHH

H

YYYYYYYYYYYYYY

YYYYYYYYYYYYYY

YYYYYYYYYYYYYY

YYY 4, 2

JJJJ

JJJJ

J

TTTTTTTT

TTTTTTTT

TT2 4,

uuuuuuuuu

kkkkkkkk

kkkkkkkk

k4, 3

ttttttttt

eeeeeeeeeeeeee

eeeeeeeeeeeeee

eeeeeeeeeeeeee

e

ZZZZZZZZZZZZZZ

ZZZZZZZZZZZZZZ

ZZZZZZZZZZZZZZ

ZZZZZ 4

XXXXXXXXXX

XXXXXXXXXX

XXXXXXXXXX

XXXX 2

TTTTTTTT

TTTTTTTT

TTTT6

FFFF

FFFF

F4

yyyyyyyy

5

kkkkkkkk

kkkkkkkk

kkk3

gggggggggg

gggggggggg

gggggggggg

gg 7

eeeeeeeeeeeeee

eeeeeeeeeeeeee

eeeeeeeeeeeeee

eee

e

Note the symmetry about the cyclic subgroups generated by powers of . This can bedescribed in terms of group theory! Indeed, there is a group isomorphism

: D8 D8defined by (n) = n and (n) = n+1. This homomorphism reflects the diagramabove about the middle line. Whenever you see symmetry you should ask yourself: whatis the group theoretical explanation?

(c) This is a simple computation:

H = {1, 2, 4, 6}, H = {, 3, 5, 7},H = {, 2, 4, 6}, H = {, 3, 5, 7}.

These cosets do in fact partition D8 as claimed, so that

D8/H = {H, H, H, H}.

(d) Lets first just double check that every nontrivial element of D8/H has order 2, to helpclarify our understanding of quotients:

(H)2 = 2H = H because 2 H;(H)2 = 2H = H because 2 = e H;

(H)2 = ()2H = H because ()2 = e H.

Now, there are several isomorphisms : D8/H (Z/2Z) (Z/2Z). For example, onecould map

(H) = (0, 0), (H) = (1, 0), (H) = (0, 1), (H) = (1, 1).

This is clearly bijective and its not hard to check that its a group homomorphism usingthe computation above.

ASSIGNMENT 4 5

(e) Since the only nontrivial subgroups of (Z/2Z) (Z/2Z) are the three cyclic ones of order2, the same is true for D8/H and we deduce that it has the subgroup lattice:

D8/H

IIII

IIII

I

vvvvvvvvv

H

HHHH

HHHH

HH H

uuuuuuuuu

H

D8

MMMMMM

MMMMM

rrrrrr

rrrrr

2,

KKKKKK

KKKK

2,

rrrrrr

rrrr

H = 2On the right we have drawn the top of the subgroup lattice for D8. Note the equalitybetween the two! This is explained by the subgroup correspondence theorem for quotientgroups.

Exercise 5 Let

Q4 = {1, a, a2, , a7, b, ba, ba2, , ba7}denote the generalized quaternion group of order 16, where a4 = b2, a8 = b4 = 1 and ab = ba7. LetH = {1, a2, a4, a6} and note that H is a normal subgroup of G.

(a) Show that Q4 contains exactly one element of order 2.(b) Use part (a) and part (a) of exercise 4 to deduce that Q4 and D8 are not isomorphic.(c) Write down the subgroup lattice for Q4.(d) Show that there is a partition

Q4 = H aH bH baH.(e) Prove that Q4/H is isomorphic with (Z/2Z)(Z/2Z). Note that as in the previous exercise,

the subgroup lattice of Q4/H is a sublattice of the subgroup lattice of Q4. (You dont haveto do anything about the subgroup lattice in this exercise, as the computation is the sameas in exercise 4. Were simply pointing out that the lattice of the quotient always appearsas a sublattice of the original groups lattice.)

Solution 5

(a) First consider an element of the form ban. Note that

(ban)2 = banban = b2a7n+n = b2(a8)n = b2 = a4 6= 1.We see that each element of the form ban is of order 4. Hence, its only possible to findelements of order two within the cyclic subgroup a of order 8, and the only element oforder 2 in there is a4. Hence Q4 contains a unique element of order 2.

(b) Since D8 has 9 elements of order 2 and Q4 only has one, these two groups are not isomorphic.(c) First note that Q4 is generated by a and any one of ba

n, for then we may cancel the an toshow that b is also in the subgroup as Q4 is generated by a and b, we get everything thisway. This is true if we pair a3, a5 and a7 with some ban too, as each of the odd powers ofa generates the cyclic subgroup a.

What happens if we pair a4 with some ban? Weve already shown in part (a) thata4 = (ban)2, so that in fact the group generated by a4 and ban is just the cyclic subgroupof order 4 generated by ban! So we get nothing new this way.

6 ASSIGNMENT 4

When we pair a2 (or a6) with the various elements ban we get interesting stuff. In thisway we find two copies of the quaternion group of order 8 inside Q4:

b, a2 = ba2, a2 = ba4, a2 = ba6, a2 = {1, a2, a4, a6, b, ba2, ba4, ba6},ba, a2 = ba3, a2 = ba5, a2 = ba7, a2 = {1, a2, a4, a6, ba, ba3, ba5, ba7}.

One can show that these are isomorphic with the quaternion group Q as follows: to seethat b, a2 = Q, map b 7 i, a2 7 j and ba2 7 k. One can treat ba, a2 similarly.

If we pair two powers an and am we will just generate a cyclic subgroup of a. So nowwe need to check what happens if we pair two elements ban and bam. If n and m are botheven then its not hard to check that you get b, a2. If n and m are both odd then youget ba, a2. Finally, if n and m have opposite parity, then you can show that the subgroupthey generate contains a, and hence must be all of Q4.

What happens if we add new elements to one of the groups generated by 2 things? Wellin this case there are only two groups that are generated by 2 elements, and theyre bothof size 8 inside Q4. By Lagrange there are no subgroups of size between 8 and 16 in Q4, sowe necessarily have to get the entire group if we add a new element to one of the groupsgenerated by 2 things. So weve found all the subgroups!

Here is the diagram:

Q4

HHHH

HHHH

H

wwwwwwwww

b, a2

FFFF

FFFF

yyyyyyyy

a ba, a2

vvvvvvvvv

IIII

IIII

I

b

RRRRRR

RRRRRR

RRRRRR

R ba2

FFFF

FFFF

a2 ba

vvvvvvvvv

ba3

jjjjjjjj

jjjjjjjj

jjjj

a4

1What subgroups are normal? Can you describe the symmetry of this diagram group theo-retically?

(d) As above, this is a simple computation:

H = {1, a2, a4, a6}, aH = {a, a3, a5, a7},bH = {b, ba2, ba4, ba6}, baH = {ba, ba3, ba5, ba7}.

These cosets do in fact partition Q4 as claimed, so that

Q4/H = {H, aH, bH, baH}.(e) As above, lets first just double check that every nontrivial element of Q4/H has order 2,

to help clarify our understanding of quotients:

(aH)2 = a2H = H because a2 H;(bH)2 = b2H = H because b2 = a4 H;

(baH)2 = (ba)2H = H because (ba)2 = a4 H.

ASSIGNMENT 4 7

Now, there are several isomorphisms : Q4/H (Z/2Z) (Z/2Z). For example, onecould map

(H) = (0, 0), (aH) = (1, 0), (bH) = (0, 1), (baH) = (1, 1).

This is clearly bijective and its not hard to check that its a group homomorphism usingthe computation above.

We said that you didnt have to do anything about the lattice, as it was similar to above,but we include a last diagram in the solutions to really drive home the point:

Q4/H

IIII

IIII

I

vvvvvvvvv

bH

HHHH

HHHH

HaH baH

uuuuuuuuu

H

Q4

LLLLLL

LLLLL

ssssss

sssss

b, a2

KKKKKK

KKKK

a ba, a2

rrrrrr

rrrr

H = a2

Remark. Exercises 4 and 5 give examples of two groups G1 and G2 of the same size which arenot isomorphic, but which have normal subgroups H1 G1, H2 G2 such that H1 = H2 andG1/H1 = G2/H2.