Assignment 10c
11, 12, 13, 37, 51, 53
11. Prove that (Z ⊕ Z)/(h(a, 0)i × h(0, b)i) is isomorphic to Za ⊕ Zb .
Note that the normal subgroup here (h(a, 0)i × h(0, b)i) is the internal direct product and can
also be written as {(na, mb) | n, m ∈ Z}.
Define φ : Z → Za ⊕ Zb by φ(x) = (x mod a, x mod b). We will show that φ is an onto
homomorphism with ker φ = {(na, mb) | n, m ∈ Z}.
Let (x, y), (z, w) ∈ Z ⊕ Z. Then
φ((x, y) + (z, w)) = φ(x + z, y + w) = (x + z mod a, y + w mod b)
= (x mod a, y mod b) + (z mod a, w mod b) = φ(x, y) + φ(z, w).
Thus φ is a homomorphism.
Let (j, k) ∈ Za ⊕ Zb . Then j, k ∈ Z with 0 ≤ j < a and 0 ≤ k < b. So (j, k) ∈ Z ⊕ Z and
φ(j, k) = (j, k). Thus φ is onto.
Suppose φ(x, y) = (0, 0). Then (x mod a, y mod b) = (0, 0) which means that x mod a = 0
and y mod b = 0. So x = na and y = mb for some n, m ∈ Z. Therefore ker φ = {(na, mb) |
n, m ∈ Z}.
Then by the First Isomorphism Theorem, (Z ⊕ Z)/(h(a, 0)i × h(0, b)i) is isomorphic to Za ⊕ Zb .
12. Suppose that k is a divisor of n. Prove that Zn /hki ≈ Zk .
Define φ : Zn → Zk by φ(x) = x mod k. This is well defined since k|n: if x ≡ y mod n then
x = y + nz for some integer z and φ(x) = x mod k = y + nz mod k = y mod k = φ(y). We
now show φ is a homomorphism, onto, and that the kernel is hki.
Let x, y ∈ Zn . Then φ(x + y) = (x + y) mod k = (x mod k) + (y mod k) = φ(x) + φ(y) so φ
is a homomorphism.
Let z ∈ Zk . Then z ∈ Z and 0 ≤ z < k so φ(z) = z and φ is onto.
If φ(x) = 0 then x mod k = 0 so x = jk for some integer j. Thus ker φ = {jk | j ∈ Z} = hki.
So by the First Isomorphism Theorem, Zn /hki ≈ Zk .
(FYI: Note that this could also be proven using the Third Isomorphism Theorem (problem 10.42),
example 13, and problem 10.51 to get Zn /hki ≈ (Z/nZ)/(kZ/nZ) ≈ Z/kZ ≈ Zk . Of course, we
used the First Isomorphism Theorem to prove the Third Isomorphism Theorem.)
13. Prove that (A ⊕ B)/(A ⊕ {e}) ≈ B.
Define φ : A ⊕ B → B by φ(a, b) = b.
Let (a, b), (c, d) ∈ A ⊕ B. Then φ((a, b)(c, d)) = φ(ac, bd) = bd = φ(a, b)φ(c, d) so φ is a
homomorphism.
Let b ∈ B. Then (e, b) ∈ A ⊕ B and φ(e, b) = b so φ is onto.
Let (a, b) ∈ A ⊕ B with φ(a, b) = e. Then b = e so (a, b) = (a, e) and ker φ = {(a, e) | a ∈ A} =
A ⊕ {e}.
So by the First Isomorphism Theorem, (A ⊕ B)/(A ⊕ {e}) ≈ B.
37. Let H = {z ∈ C∗ | |z| = 1}. Prove that C∗ /H is isomorphic to R+ , the group of positive real
numbers under multiplication.
Define φ : C∗ → R+ by φ(z) = |z|.
Let z, w ∈ C∗ . Then φ(zw) = |zw| = |z||w| = φ(z)φ(w). So φ is a homomorphism.
Let r ∈ R+ . Then r ∈ C∗ and φ(r) = |r| = r so φ is onto.
Let z ∈ C∗ such that φ(z) = 1. Then |z| = 1. So ker φ = S 1 = {z | |z| = 1}.
Therefore, by the First Isomorphism Theorem we see that C∗ /H ≈ R+ .
∗
iθ
(FYI: You can
) = r or
√ also use other notations for C or the function φ, for example φ(re
2
2
2
φ(a + bi) = a + b . You can also use the slightly different function φ̃(a + bi) = a + b2 .)
51. Let N be a normal subgroup of a group G. Use property 7 of Theorem 10.2 to prove that every
subgroup of G/N has the form H/N , where H is a subgroup of G.
Consider the homomorphism φ : G → G/N defined by φ(g) = gN (as in Theorem 10.4). Let
K ≤ G/N . Then by 10.2.7 we know that φ−1 (K) is a subgroup of G, let’s call it H. So
H = φ−1 (K).
Next we show that φ is onto: Let gN ∈ G/N . Then g ∈ G and φ(g) = gN . So φ is onto which
means then that φ(H) = K. (If we didn’t know that φ was onto, then we would only have a
subset here, not an equality.)
So K = φ(H) = {φ(h) | h ∈ H} = {hN | h ∈ H} = H/N , where H is a subgroup of G.
53. Use the First Isomorphism Theorem to prove Theorem 9.4:
For any group G, G/Z(G) is isomorphic to Inn(G).
Define ψ : G → Inn(G) by ψ(g) = φg . Recall that φg is the inner automorphism induced by g:
φg (x) = gxg −1 .
Let g, h ∈ G. Then ψ(gh) = φgh = φg φh = ψ(g)ψ(h) by problem 11 in chapter 6. (Recall that
the operation in Inn(G) is composition.) So ψ is a homomorphism.
Let φg ∈ Inn(G). Then g ∈ G and ψ(g) = φg so ψ is onto.
Note that the identity in the Inner Automorphism group is the identity function: φe (x) =
exe−1 = x. Suppose ψ(a) = φe . Then φa = φe which means that for every x ∈ G, φa (x) = φe (x)
or axa−1 = x. This is equivalent to ax = xa for every x ∈ G, thus a ∈ Z(G). Therefore
ker ψ = Z(G).
Then the First Isomorphism Theorem tells us that G/Z(G) ≈ Inn(G).