Homework 03 - Ma 5a - Solutions
Luciena Xiao
October 27, 2016
1.(a) (5 points) Define φ : G/K → G/H as follows: for any g ∈ G, φ(gK) = gH. We first check
that φ is well-defined: since K is a subgroup of H, g1 K = g2 K ⇒ g1 = g2 k for some k ∈ K ⇒
g1 H = g2 kH ⊂ g2 H. By problem 3(c) from homework set 2, we get g1 H = g2 H. To see that φ is
surjective, note that for any coset gH of H, the coset gK of K is mapped to gH under φ.
Define ψ : H/K → G/K to be the identity map, i.e. ψ(gK) = gK for any g ∈ H. ψ is well-defined
because cosets do not depend on choice of representatives. φ is tautologically injective.
1.(b) (5 points) By the existence of a surjective map from G/K to G/H and the definition of
index, we have [G : H] ≤ [G : K]. Similarly, the existence of an injective map from H/K to G/K
implies [H : K] ≤ [G : K]. Thus [G : H] and [H : K] are both finite when [G : K] is finite.
1.(c) (5 points) Suppose [G : H] = s, [H : K] = r. Let g1 , · · · , gs ∈ G be a list of coset representatives of H in G. Let h1 , · · · , hr be a list of coset representatives of K in H. Note that
G = ∪si=1 gi H = ∪si=1 (gi ∪rj=1 hj K) = ∪si=1 ∪rj=1 gi hj K, so [G : K] ≤ [G : H][H : K]. To prove equality, it suffices to show that for gi hj K 6= gm hn K whenever gi 6= gm , hj 6= hn . But gi hj K = gm hn K ⇒
−1 −1
−1
−1
hj K = gi−1 gm hn K ⇒ K = h−1
j gi gm hn K ⇒ hj gi gm hn ∈ K ⊂ H ⇒ gi gm ∈ H, where the last
statement is only possible when gi−1 gm = e, so gi = gm , which then implies hj K = hn K, which is
only possible when hj = hn . Thus we conclude [G : K] = [G : H][H : K].
Remark. In this problem, one cannot use [G : K] = |G|/|H| etc., because the expression on the
right hand side only makes sense when |G| is finite.
2.(b) (5 points) If gHg −1 ⊂ H for all g ∈ G, then (g −1 )−1 Hg −1 ⊂ H for all g ∈ G. As g −1 ranges
over all elements of G, we obtain g −1 Hg ⊂ H for all g ∈ G, which implies H ⊂ gHg −1 for all
g ∈ G, so we conclude H = gHg −1 .
2.(c) (5 points) First suppose |H| is finite. Note that conjugation by g is injective on G, so in
particular it is injective on H; therefore |gHg −1 | = |H|, which allows us to conclude gHg −1 = H.
Now suppose [G : H] is finite, then H has finitely many cosets, so we must have H = g n H for some
positive integer n, which implies g n ∈ H. Since conjugation by an element of H is an automorphism
on H, we have H = g n H(g n )−1 = g n Hg −n = g n−1 (gHg −1 )g −(n−1) ⊂ g n−1 Hg −(n−1) ⊂ · · · ⊂
gHg −1 ⊂ H. Thus we conclude H = gHg −1 .
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3.(a) (5 points) Let g ∈ G be any element. If g ∈ H, then gH = H = Hg. Otherwise, gH 6= H
and Hg 6= H. But H has index 2, so we must have gH = G − H = Hg. Thus H is normal.
3.(b) (5 points) Write τ = (a1 · · · am ). Let x ∈ {1, · · · , n}. If σ −1 (x) ∈
/ {a1 , · · · , am }, then σ ◦ τ ◦
−1
−1
−1
σ (x) = σ ◦ σ (x) = x. If σ (x) = ai for some 1 ≤ i ≤ m, then σ ◦ τ ◦ σ −1 (x) = σ ◦ τ (ai ) =
σ(ai+1 mod m ). Thus we obtain the desired equality.
3.(c) (5 points) By Lagrange’s theorem, the subgroups of S3 must have order 1, 2, 3, 6. There is
only one possible subgroup of order 1, namely {e}, which is obviously normal since e commutes
with any element of S3 . There is only one possible subgroup of order 6: S3 , which is trivially
normal. Each subgroup of order 2 is generated by an element of order 2, so such subgroups are
h(12)i, h(23)i, h(13)i, among which none is normal because (12)(23) = (123) 6= (132) = (23)(12)
and (12)(13) = (132) 6= (123) = (13)(12). Lastly, each subgroup of order 3 is generated by an
element of order 2, but h(123)i = h(132)i, so there is only one such subgroup; moreover, this
subgroup has index 2, which by part (a) is normal. To summarize, the normal subgroups of S3 are
{e}, S3 , {e, (123), (132)}.
4.(a) (10 points) First note that by Lagrange’s theorem, a subgroup of index n has order 2n/n = 2,
so each subgroup of this kind is generated by an element of order 2. In D2n , the elements of order
2 are of the form sj ri , 0 ≤ i ≤ n, 0 ≤ j ≤ 1 and rn/2 if n is even. For the subgroup to be normal,
the image of the nontrivial element under conjugation by any element of D2n must be itself. But
rsj ri r−1 = sj r−1 ri rn−1 = sj ri+n−2 , which is equal to sj ri only when i = i + n − 2, which is
impossible when n ≥ 5. Thus D2n has no normal subgroup when n is odd.
Now, when n is even, to see that hrn/2 i is normal, it suffices to show srn/2 s−1 = rn/2 and rrn/2 r−1 =
rn/2 as s and r generate D2n , and one can check by simple computations that these equalities indeed
hold.
5.(a) (5 points) Recall that it is showed in class that conjugation by any element of G is an
automorphism of G, so by the definition of a characteristic group, gHg −1 = H for all g, which
implies H is normal.
5.(c) (5 points) Let g ∈ G be any element and let φg denote the conjugation by g. Then φg (H) = H
since H is normal in G. Thus φg |H is an automorphism of H, which implies φg |H (K) = K since K
is characteristic in H. Hence φg (K) = K for all g ∈ G, i.e. K is normal in G.
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