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Math 210A: Algebra, Homework 2
Ian Coley
October 17, 2013
Problem 1.
Let G be a group and a, b ∈ G.
(a) Prove that an · · · am = an+m and (an )m = anm .
(b) Prove that ord an =
ord a
(n,ord a)
if ord a < ∞.
(c) Prove that ord ab = ord a · ord b if a and b commute and (ord a, ord b) = 1.
Solution.
(a) We proceed by induction on m. It is clear that an · a = an+1 and an · a−1 = an−1 . The
case where m is positive is analogous to the case where m is negative, so we prove it
only in the positive case. Suppose the case m = k. Then
an · ak+1 = an · ak · a = an+k · a = an+k+1 .
Similarly, it is clear that (an )1 = an = a1·n . Assume the case m = k. Then
(an )k+1 = (an )k · an = ank · an = ank+n = a(k+1)n .
where the penultimate inequality comes from the previous part of this question. This
completes the proof.
(b) First, let k =
ord a
.
(n,ord a)
Since (n, ord a) | n, write d · (n, ord a) = n. Then
(an )k = a(n·ord a)/(n,ord a) = ad·ord a = (aord a )d = e.
Suppose now that (an )` = e. Then by the division algorithm, write ` = qk + r, where
0 ≥ r < k. Then
(an )` = (an )qk+r = (an )r = arn
Therefore we need arn = e, so ord a | rn. In particular, lcm(n, ord a) | rn and
n·ord a
lcm(n, ord a) = (n,ord
. Then for some d0 ∈ N,
a)
rn = d0 ·
n · ord a
ord a
⇐⇒ r = d0
= d0 k.
(n, ord a)
(n, ord a)
Since r < k by assumption, this implies that d0 = 0 so r = 0. Therefore ` = qk, so k is
indeed the order of an .
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(c) Let ord a = m and ord b = n such that (m, n) = 1. We see that (ab)mn = amn bmn =
(am )n (bn )m = e, where we use commutativity in the second equality. Now suppose
(ab)k = e. By the division algorithm, we can write k = qm + r, where 0 ≥ r < m.
Then
(ab)k = (ab)qm+r = aqm ar · bqm+r = ar bqm+r ,
so we must have r = 0. Therefore m | k. Similarly, if we write k = q 0 n + r0 , where
0 ≥ r < n, we have
0
0
0
0
0
(ab)k = (ab)q n+r = br aq n+r .
So r0 = 0 and n | k. Therefore lcm(m, n) | k. In general we have lcm(m, n) =
mn/(m, n). But (m, n) = 1, so mn | k. Therefore k = mn is the smallest number such
that (ab)k = e, and we are done.
Problem 2.
Let H ⊂ G be a subgroup. Show that the correspondence Ha 7→ a−1 H is a bijection between
the sets of right and left cosets.
Solution.
Let ϕ be this correspondence. Suppose ϕ(Ha) = ϕ(Hb). Then a−1 H = b−1 H, so ba−1 H =
H, which implies ba−1 ∈ H. This is equivalent to saying that Ha = Hb. Thus ϕ is injective.
It is clear that ϕ is surjective since the number of left and right costs is equal. This ϕ is a
bijection.
Problem 3.
Let H ⊂ G be a subgroup. Suppose that for any a ∈ G there exists b ∈ G such that
aH = Hb. Show that H is normal in G.
Solution.
Let a = ae ∈ aH. Then a ∈ Hb by assumption. Additionally, a = ea ∈ Ha, so a ⊂ Ha ∩ Hb.
Since cosets are either disjoint or identical, this implies Ha = Hb. Therefore aH = Ha for
all a ∈ G, so H is normal in G.
Problem 4.
Let f : G → H be a surjective group homomorphism.
(a) Let H 0 be a subgroup of H. Show that G0 = f −1 H 0 is a subgroup of G. Prove that the
correspondence H 0 7→ G0 is a bijection between the set of all subgroups of H and the
set of all subgroups of G containing ker f .
(b) Let H 0 be a normal subgroup of H. Show that G0 = f −1 H 0 is a normal subgroup of G.
Prove that G/G0 ∼
= H/H 0 and the correspondence H 0 7→ G0 is a bijection between the
set of all normal subgroups of H and the set of all normal subgroups of G containing
ker f .
Solution.
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(a) Clearly eG ∈ G0 since eG ∈ f −1 (eH ) and eH ∈ H 0 . Let a, b ∈ G0 . Then let f (a) = x ∈ H 0
and f (b) = y ∈ H 0 . By assumption, f (a)f (b) = f (ab) = xy ∈ H 0 since it is a
subgroup, so ab ∈ G0 . If a ∈ G0 and f (a) = x as above, then x−1 ∈ H 0 . Since
f (a−1 ) = f (a)−1 = x−1 , we have a−1 ∈ G0 . This shows that G0 is a subgroup.
Since f is surjective, we know that G/ ker f ∼
= H. Therefore any subgroup of G/ ker f ,
i.e. a subgroup which contains ker f , maps isomorphically to a subgroup of H. Similarly, any subgroup of H pulls back to a subgroup of G/ ker f , which is a subgroup of
G containing the kernel, and it is clear that these are mutually inverse since f is an
isomorphism on G/ ker f → H. This completes the proof.
(b) We know that G0 is a subgroup of G by the above. Further, by last week’s assignment,
the preimage of any normal subgroup in H is normal in G, and further it must contain
ker f . We need only show that the image of a normal subgroup of G/ ker f is normal
in H, and we will be done by identical reasoning as the above. Let G0 be such a
subgroup, and let H 0 be its image, which is a subgroup. Then since xgx−1 ∈ G0 for
every g ∈ G0 , x ∈ G/ ker f , we have
ϕ(g) ∈ H and ϕ(xgx−1 ) = ϕ(x)ϕ(g)ϕ(x)−1 ∈ H.
Since ϕ(x) runs through every element of H and ϕ(g) every element of H 0 , H 0 is
preserved under conjugation, i.e. it is normal. Since f is an isomorphism of G/ ker f
and H, this correspondence is bijective, and we are done.
Problem 5.
(a) Let N be a subgroup in the centre Z(G) of G. Show that N is normal in G. Prove
that if the factor group G/N is cyclic, then G is abelian.
(b) Prove that every group of order p2 (for a prime p) is abelian.
Solution.
(a) We use the definition N / G if and only if xN x−1 ⊂ N for all x ∈ G. Let g ∈ N . Then
xgx−1 = xx−1 g = g, since g ∈ Z(G), and hence xgx−1 ∈ N for all x ∈ G, g ∈ N .
Hence N is normal.
We claim that if G/N is cyclic that G is abelian. Write G/N = hxN i. Now let a, b ∈ G
such that a ∈ (xZ(G))m = xm Z(G) and b ∈ xn Z(G). Then a = xm g1 and b = xn g2 for
g1 , g2 ∈ N . Then
ab = (xm g1 )(xn g2 ) = xm (g1 xn )g2 = xm (xn g1 )g2 = (xm xn )(g1 g2 ) = (xn xm )(g2 g1 )
= xn (xm g2 )(g1 ) = xn (g2 xm )g1 = (xn g2 )(xm g1 ) = ba.
We use the fact that g1 , g2 ∈ Z(G) and that x commutes with itself freely. This shows
that G is abelian.
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(b) Let G act on itself by conjugation. Then by the class equation
X
p2 = |G| = |Z(G)| +
|G : NG (g)|.
nontrivial conjugacy classes
By Lagrange’s theorem, |G : NG (g)| = |G|/|NG (g)| and since these conjugacy classes
are proper and nontrivial, p | |G : NG (g)|. Since p | p2 , we must have p | |Z(G)|, so
Z(G) is nontrivial. If |Z(G)| = p2 , we are done.
Suppose |Z(G)| = p. By 5(a), Z(G) is normal so we may examine the factor group
G/Z(G). Since |G/Z(G)| = |G|/|Z(G)| = p, by earlier considerations G/Z(G) must be
the cyclic group of order p, which implies that G is abelian, contradicting |Z(G)| = p.
Therefore Z(G) = G and we are done.
Problem 6.
Prove that if a group G contains a subgroup H of finite index, then G contains a normal
subgroup of finite index.
Solution.
Let |G : H| = n < ∞. Taking the hint, examine the action of G onto the finite set of n
cosets aH by left multiplication. This induces a homomorphism ϕ : G → Sn . By earlier
considerations, ker ϕ is a normal subgroup of G. We claim that it has finite index. By the
first isomorphism theorem, G/ ker ϕ ∼
= H for some subgroup H ⊂ Sn . Thus |G/ ker ϕ| < ∞
since Sn is a finite group. But |G : ker ϕ| = |G/ ker ϕ| < ∞, thus ker ϕ / G is a normal
subgroup of finite index.
Problem 7.
(a) Show that the group Inn G of all inner automorphisms of a group G is a normal
subgroup in Aut G.
(b) Find all automorphisms of all (finite and infinite) cyclic groups.
Solution.
(a) Let fg be the inner automorphism given by fg (a) = gag −1 and let ϕ be any automorphism of g. Then for a ∈ G,
(ϕ ◦ fg ◦ ϕ−1 )(a)ϕ ◦ fg (ϕ−1 (a)) = ϕ(gϕ−1 ag −1 )
= ϕ(g)ϕ(ϕ−1 (a))ϕ(g −1 ) = ϕ(g)aϕ(g)−1 ,
where ϕ(ϕ−1 (a)) = a because ϕ is an automorphism, hence bijective. Therefore ϕ ◦
fg ϕ−1 = fϕ(g) , so inner automorphisms are preserved under conjugation by an arbitrary
automorphism. thus Inn G / Aut G.
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(b) Let G be a cyclic group, and let G× be the multiplicative group of G, i.e. its set of
generators. From previous considerations, if |G| = n, |G| = ϕ(n) where ϕ is Euler’s
totient function. If G = Z, the only infinite cyclic group up to isomorphism, then G× =
{−1, 1} so |G| = 2. We claim that automorphisms of G are in bijective correspondence
with elements of G× .
Suppose ϕ is an automorphisms of G = hgi, where g is a fixed generator. Then ϕ(g)
determines the entire map. If a = g m , then ϕ(a) = ϕ(g m ) = ϕ(g)m . Therefore if ϕ is
an automorphism, then ϕ(g) must generate G. Conversely, if h is another generator
for G, then the map g 7→ h is an automorphism. Hence every automorphism of a cyclic
group G = hgi is of the form g 7→ h ∈ G× .
Problem 8.
Prove that if G has no non-trivial automorphisms, then G is abelian and g 2 = e for all g ∈ G.
Solution.
Let ϕg be the automorphism of G given by conjugation by g. Then by assumption ϕg = 1,
the identity automorphism. Therefore ϕg (a) = gag −1 = a for all a, g ∈ G. This implies that
ga = ag for all a, g ∈ G, i.e. G is abelian.
Let ϕ : G → G given by ϕ(g) = g −1 . We claim this is an automorphism of G. ϕ(g)ϕ(h) =
g −1 h−1 , and since G is abelian, we have g −1 h−1 = h−1 g −1 = (gh)−1 = ϕ(gh), so ϕ preserves
multiplication. Clearly ker ϕ = e since g −1 = e ⇐⇒ g = e. Further, ϕ is surjective since
ϕ(g −1 ) = g for all g ∈ G. Hence G is an automorphism, and must be the identity. Therefore
ϕ(g) = g −1 = g for all g, and left multiplication by g shows e = g 2 for all g ∈ G. This
completes the proof.
Problem 9.
Let x and x0 be two elements in the same orbit under some action of a group G on a set.
Show that the stabilisers Gx and Gx0 are conjugate in X.
Solution.
Suppose that x0 = gx. By definition we also have g −1 x0 = x. Let h ∈ Gx . Then
ghg −1 x0 = ghg −1 (gx) = ghx = gx = x0 .
Therefore gGx g −1 ⊂ Gx0 . Conversely, let h0 ∈ Gx0 . Then
g −1 h0 gx = g −1 h0 x0 = g −1 x0 = x.
Hence g −1 Gx0 g ⊂ Gx ⇐⇒ Gx0 ⊂ gGx g −1 . Since we have shown double inclusion, we have
Gx0 = gGx g −1 , i.e. the stabilisers of the x and x0 are conjugate.
Problem 10.
Let a group G act on two sets X and Y . We say that X and Y are G-isomorphic if there is
a bijection f : X → Y such that f (gx) = g(f (x)) for every x ∈ X and g ∈ G. Prove that
if G acts on X transitively, then X is G-isomorphic to the set of left cosets G/H for some
subgroup H ⊂ G (with the action of G on G/H by left translation).
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Solution.
Since G acts on X transitively, this action has only one orbit, namely G itself. Let Gx be
the stabiliser of some x ∈ G. We claim that f : X → G/Gx is a G-isomorphism, where
f (y) = hGx , where hx = y. First, we show this is well-defined. Suppose that hx = h0 x = y.
Then h−1 h0 x = x, so h−1 h0 ∈ Gx . Therefore hGx and h0 Gx represent the same coset.
By the orbit-stabiliser theorem, we know |Gx| = |G|/|Gx | for all x ∈ X. Since the
action is transitive, Gx = X, so |Gx| = |X|. Further, the number of cosets G/Gx is equal
to |G : Gx | = |G|/|Gx |. Therefore f is a map between sets of equal cardinality. Suppose
f (y) = f (y 0 ) = hGx . Then hx = y = y 0 . Therefore f is injective, and since |X| = |G/Gx |, f
is bijective.
Finally we prove that f is a G-isomorphism. Let gy = z, hx = z, and kx = y. Then
f (gy) = f (z) = hGx , and g(f (y)) = gkGx . Then by construction, hx = gkx = z, so
h−1 gk ∈ Gx , thus hGx and gkGx represent the same coset. This proves that f is a Gisomorphism, so we are done.
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