Midterm review solutions
These are a few problems to help you prepare for the midterm. You don’t have to turn them in.
1. (a) First, note that ρh (n) ∈ N for all h ∈ H and n ∈ N , because N is normal. We claim that
ρh : N → N is an automorphism for all h ∈ H. It is a homomorphism, because
(ρg ◦ ρh )(x) = h−1 g −1 xgh = ρgh (x)
for all g, h ∈ H. It is invertible, because ρh ◦ ρh−1 (x) = ρe (x) = x and ρh−1 (x) ◦ ρh = ρe (x) = x.
Therefore, it is an automorphism.
Next, we show that ρ : H → Aut(N ) is a homomorphism. If g, h ∈ H and n ∈ N , then
(ρg ◦ ρh )(n) = ρh (ρg (n))
= ρh (g −1 ng)
= h−1 g −1 ngh = ρgh (n),
so ρg ◦ ρh = ρgh . This implies that ρ is a homomorphism.
∼ H nρ N .
(b) Show that G =
Recall that H nρ N is the group whose elements consist of the products of the form hn for h ∈ H
and n ∈ N , and whose multiplication is given by
(hn)(h0 n0 ) = (hh0 )(ρh0 (n)).
From the last assignment, we know that G is the set of products of the form hn for h ∈ H and
n ∈ N . We need to show that the multiplication forumla above holds in G. But
(hn)(h0 n0 ) = h(h0 (h0 )−1 )nh0 n0 = hh0 ρh0 (n)n0 ,
so G ∼
= H nρ N .
∼ Z10 .
2. (a) Show that Aut(Z11 ) =
Since Z11 is cyclic, a homomorphism is determined by where it sends 1 – if f is a homomorphism
and f (1) = n, then we must have f (x) ≡ nx (mod 11). Therefore the set of homomorphisms
from Z11 to itself is the set {f0 , . . . , f10 }, where fn (x) ≡ nx (mod 11); there is a homomorphism
taking 1 to each element of Z11 .
The map fn is an automorphism if and only if n is relatively prime to 11, so all of these are
automorphisms except f0 . Aut(Z11 is therefore a group with 10 elements. We claim it is cyclic.
But a simple calculation shows that the powers of f2 are
(f2 , f22 , . . . , f211 ) = (f2 , f4 , f8 , f5 , f10 , f9 , f7 , f3 , f6 , f1 ),
so Aut(Z11 ) is the cyclic group of order 10, i.e., Z10 .
(b) A semidirect product of Z5 and Z11 would have 55 elements. We therefore look for a homomorphism ρ : Z5 → Aut(Z11 ). Let ρ be the homomorphism that sends 1 to f4 ; by the calculation
above, (f4 )5 = f1 = id, so there is such a homomorphism. Let G = Z11 nρ Z5 . Then, if we write
elements of G as pairs (m, n) with m ∈ Z11 and n ∈ Z5 , we can check that (1, 0)(0, 1) = (1, 1),
but (0, 1)(1, 0) = (1, f4 (1)) = (1, 4), so the group is not abelian.
3. (a) Show that {−1, 1} is a normal subgroup of Q. What are the elements of Q/{−1, 1}? Show that
Q/{−1, 1} is isomorphic to Z2 × Z2 .
In fact, for every x ∈ Q, we have (−1)x = x(−1) = −x, so {−1, 1} is a normal subgroup. Its
cosets are
C1 = {−1, 1}
Ci = {−i, i}
1
Cj = {−j, j}
Ck = {−k, k}.
Since i2 = j 2 = k 2 = −1, we have
Ci2 = Cj2 = Ck2 = C1 .
By similar calculations, we can construct the following multiplication table:
×
C1
Ci
Cj
Ck
C1
C1
Ci
Cj
Ck
Ci
Ci
C1
Ck
Cj
Cj
Cj
Ck
C1
Ci
Ck
Ck
Cj
Ci
C1
This is exactly the multiplication table of Z2 × Z2 , so the two groups are isomorphic.
(b) By Lagrange’s Theorem, the subgroups must have order 1, 2, 4, or 8.
Order 1 Only the trivial subgroup
Order 2 Subgroups like this must consist of the identity and an element of order 2. The elements
±i, ±j, and ±k all have order 4, so the only order 2 subgroup is {−1, 1}
Order 4 There are three cyclic subgroups of order 4, generated by i, j, or k. We claim that these
are the only ones. Suppose that H is a subgroup of Q of order 4 that is not one of the groups
above. Then it can’t contain i because then it would contain the cyclic subgroup generated
by i, which already has four elements in it. Likewise, it can’t contain ±i, ±j, or ±k. All
that’s left is ±1, so there aren’t enough elements to form another subgroup of order 4.
Order 8 Only the whole group.
Since all the order-4 subgroups are cyclic, none of them are isomorphic to Z2 × Z2 .
4. If Q is a quotient group of G, then Q = G/N for some normal subgroup N . Then, for any a, b ∈ G,
we have
(N a)(N b) = N ab = N ba = (N b)(N a),
so Q is abelian.
5. Suppose that f : G → H is a homomorphism and a ∈ G. If o(a) = 15, what are the possible values of
o(f (a))?
We know that a15 = e, so f (a)15 = f (a15 ) = e. Therefore, o(f (a)) is a divisor of 15, either 1, 3, 5, or
15. All of these are possible when G = H = Z15 and a = 1. For instance, the map f : Z15 → Z15 given
by f (x) = 3x mod 15 is a homomorphism that takes a to 3, and o(3) = 5.
6. Let G be a group and let A, B ⊂ G be normal subgroups such that A ∩ B = {e}. Show that G is
isomorphic to a subgroup of G/A × G/B.
We want to construct an injective homomorphism from G to G/A × G/B. Define
f (g) = (Ag, Bg)
Then f (gh) = (Agh, Bgh) = (Ag, Bg)(Ah, Bh), so this is a homomorphism. Its kernel is
ker(f ) = {g ∈ G | Ag = A, Bg = B}.
But Ag = A if and only if g ∈ A and Bg = B if and only if g ∈ B, so ker(f ) = A ∩ B = {e}.
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