Midterm solutions
1. (a) What is a homomorphism?
A homomorphism between two groups G and H is a map f : G → H such that f (g1 g2 ) =
f (g1 )f (g2 ) for every g1 , g2 ∈ G.
(b) Suppose that G is abelian and that f : G → H is a surjective (onto) homomorphism.
Show that H is abelian.
Suppose h1 , h2 ∈ H. Since f is surjective, there are g1 , g2 ∈ G such that f (g1 ) = h1 and
f (g2 ) = h2 . Then
h1 h2 = f (g1 )f (g2 )
= f (g1 g2 )
= f (g2 g1 )
= h2 h1
so H is abelian.
2. (a) What is a quotient group? (Be sure to describe both the set and the operation.)
If G is a group and N is a normal subgroup, then the quotient group of G by N is the
set G/N of right cosets of N in G with the operation (N a)(N b) = N ab. (Alternatively,
the operation is multiplication of sets:
C1 C2 = {c1 c2 | c1 ∈ C1 , c2 ∈ C2 }.
(b) Describe all of the quotient groups of the symmetric group S3 .
The subgroups of S3 (which is isomorphic to Sym(∆)) are:
• {e}: Normal, and the quotient group is S3 /{e} ∼
= S3
• {e, (123), (321)}: Normal (this is the alternating subgroup), and the quotient group
is S3 /{e, (123), (321)} ∼
= Z2
• {e, (12)}, {e, (23)}, {e, (13)}: None of these are normal (you can check that their
left and right cosets are different)
• S3 itself: Normal, and the quotient group is S3 /S3 ∼
= {e}
3. Suppose that A and B are groups and that f : A → B is a homomorphism. Show that the
“graph”
G = {(a, f (a)) ∈ A × B | a ∈ A}
is a subgroup of A × B and that [A × B : G] = o(B). Your proof should still hold when A is
an infinite group.
First, we show that G is a group. If (a, f (a)), (a0 , f (a0 )) ∈ G, then
(a, f (a))(a0 , f (a0 )) = (aa0 , f (a)f (a0 )) = (aa0 , f (aa0 ))
which is an element of G, so G is closed under the group operation. If (a, f (a)) ∈ G, then
(a, f (a))−1 = (a−1 , f (a)−1 ) = (a−1 , f (a−1 ))
which is an element of G, so G is closed under inverses. This makes G a subgroup.
Now, we calculate the index of G. For each b ∈ B, there is a right coset
G ◦ (e, b) = {(a, f (a)b) ∈ A × B | a ∈ A}.
These are all distinct because if G ◦ (e, b) = G ◦ (e, b0 ), then (e, b0 )(e, b)−1 = (e, b0 b−1 ) ∈ G,
which implies that b0 b−1 = e. On the other hand, if (a, b) ∈ G, then
(a, b) = (a, f (a)f (a)−1 b) ∈ G ◦ (e, f (a)−1 b),
so the union of these sets is all of G. Therefore, these are all the right cosets of G. There is
one for each element of B, so [A × B : G] = o(B).
4. Suppose that A is a finite abelian group of order n whose elements are a1 , a2 , . . . , an . Show
that (a1 a2 . . . an )2 is the identity element.
Since A is abelian, we can rearrange the product
(a1 a2 . . . an )2 = (a1 a2 . . . an )(a1 a2 . . . an )
so that each element in the first half is placed next to its inverse, coming from the second
half:
−1
(a1 a2 . . . an )2 = a1 a−1
1 . . . an an .
Each element cancels with its inverse, so the product is the identity.
(This doesn’t work for a1 a2 . . . an because some elements might be equal to their inverses.)
5. Show that Aut(Z7 ) ∼
= Z6 . Use semi-direct products to construct a non-abelian group of order
21.
Any homomorphism f : Z7 → Z7 is of the form f (x) = nx, where n = f (1). Therefore,
there are seven homomorphisms Z7 → Z7 , {f0 , . . . , f6 }, where fn (x) = nx. All of these are
invertible except f0 , so Aut(Z7 ) = {f1 , . . . , f6 }. This is cyclic because every element is a
power of f3 :
(f3 , f32 , . . . , f36 ) = (f3 , f2 , f6 , f4 , f5 , f1 ),
so it is isomorphic to Z6 .
Let ρ : Z3 → Aut(Z7 ) be the homomorphism that sends 1 to f2 . Since (f2 )3 = id, this is a
homomorphism. Let G = Z7 nρ Z3 . Then, if we write elements of G as pairs (m, n) with
m ∈ Z7 and n ∈ Z3 , we can check that (1, 0)(0, 1) = (1, 1), but (0, 1)(1, 0) = (1, f2 (1)) = (1, 2),
so the group is not abelian.
6. Find and fix the mistake in the following proof.
Theorem 1. If H is a subgroup of G, then there is a one-to-one correspondence between the
left cosets of H in G and the right cosets of H in G.
Proof. Let L = {gH | g ∈ G} be the set of left cosets and let R = {Hg | g ∈ G} be the
set of right cosets. Let f : L → R be the map f (gH) = Hg. This is invertible with inverse
f −1 (Hg) = gH, so it is a one-to-one correspondence between the left cosets of H and the
right cosets of H
The problem is that the map f (gH) = Hg isn’t well-defined. Specifically, if h ∈ H, then
gH = ghH, but f (gH) = Hg 6= Hgh = f (ghH).
In order to get a well-defined map from L to R, we can let f (gH) = Hg −1 . Then if gH = g 0 H,
then (g 0 )−1 g ∈ H, so ((g 0 )−1 g)−1 = g −1 g 0 ∈ H, so Hg −1 = H(g 0 )−1 and the map is welldefined. Since it has an inverse f −1 (Hg) = g −1 H, it is a one-to-one and onto map from L to
R.