Math 330 Review Problems for Test 2 Solutions 1. Let a and b be ...

Math 330 Review Problems for Test 2 Solutions
Proof. Let H ≤ G be a normal subgroup of G such that |G : H| = 24 and |H| = 11. Let
x ∈ G be an element such that x 11 = e. As H is normal in G, G/H is a factor group of order
|G : H| = 24. Consider the coset xH ∈ G/H. Since |G/H| = 24, (xH) |G/H| = (xH) 24 =
(x 24 )H must be the identity element by Corollary 4 to Lagrange’s Theorem. Hence x 24 ∈ H,
as x 24 H = eH. Yet x 24 = x 22 · x 2 = (x 11 ) 2 · x 2 = e 2 · x 2 = x 2 , so x 2 ∈ H. As x 2 ∈ H,
x 2n ∈ H for all n ∈ Z. In particular, x 12 ∈ H. But we already know that x 11 = e, so
x = x · e = x · x 11 = x 12 ∈ H implies that x ∈ H, which is what we needed to prove.
11. If H is a normal subgroup of G and K is any subgroup of G, prove that the subgroup H ∩ K
is normal in K.
Proof. We will apply the Normal Subgroup Test to N = H ∩ K, viewed as a subgroup of K
(we know that N ≤ K from previous work). Suppose x ∈ K and y ∈ N. Since N = H ∩ K,
y ∈ N implies that y ∈ H and y ∈ K. Thus, since K is a subgroup of G, the product
xyx −1 ∈ K since both x, y ∈ K. Since H is normal in G, the product xyx −1 ∈ H as y ∈ H
and x ∈ K ≤ G. Therefore, xyx −1 ∈ H ∩ K = N, so xNx −1 ⊆ N. Hence N is normal in K
since the element x ∈ K was arbitrary.
12. Let G be a group with more than one element. Is it ever the case that the set of all homomorphisms
from G to G is a group under function composition? Justify your answer.
Solution: No.
Proof. First of all, under function composition, the “identity element” must be the identity
function, id G : G → G given by id G (g) = g for all g ∈ G.
We know that for any pair of groups G 1 and G 2 , there is always the trivial homomorphism
ε : G 1 → G 2 defined by ε(g) = e 2 for all g ∈ G 1 , where e 2 is the identity element of G 2 . We
claim that the homomorphism ε : G → G is not invertible in the set of all homomorphisms
from G to G under function composition. For, suppose to the contrary that there is a
homorphism ϕ : G → G such that ϕ ◦ ε = id G . Let g ∈ G be an element other that e,
which exists since we are assuming that G is a group with more than one element. Consider
ϕ(ε(g)) = ϕ(e) = e ≠ g = id G (g), so ϕ ◦ ε ≠ id G . Hence the homomorphism ε has no inverse,
so the set of all homomorphisms from G to G under function composition is not a group.
13. Show that there are infinitely many homomorphisms from Z to itself.
Solution: For each integer n ≥ 0, define the function ϕ n : Z → Z by ϕ n (a) = na. Then ϕ n
is a function from Z to itself which is operation-preserving, since
ϕ n (a + b) = n(a + b) = na + nb = ϕ n (a) + ϕ n (b).
Hence, ϕ n is a group homomorphism from Z to itself for each integer n ≥ 0. We claim that
if n and m are distinct non-negative integers, then ϕ n and ϕ m are distinct homomorphisms.
To see this, it is enough to compare ϕ n (1) = n1 = n with ϕ m (1) = m1 = m. As n ≠ m,
ϕ n (1) ≠ ϕ m (1), so ϕ n ≠ ϕ m . Thus there are infinitely many group homomorphisms from Z
to itself.
14. Given that m divides n, show that Z/mZ is a homomorphic image of Z/nZ.