Solutions to Chapter 2
2.1 Let \G\ = pa. The class equation gives
where g'x G Cx for \CX\ > 1. Now p divides both \G\ and \G : M{g'x)l so
p divides |Z(G)|. Hence Z(G) is non-trivial.
Now suppose that \G\ = p2. Since Z[G) is non-trivial we have
\Z{G)\ =pov \Z{G)\ = p2. Now if |Z(G)| = p then |G/Z(G)| = p so is
cyclic. Let aZ(G) be a generator of G/Z(G). Then two arbitrary ele-
ments of G are a m x and any where x,y G Z(G). But amxany = anyamx
since re, t/ £ Z(G). Hence G is abelian, and so \G\ = \Z(G)\ — p which is
a contradiction. It follows that we must have |#(G)| = p2 — \G\ whence
G is abelian.
The groups of order 9 are C3 x C 3 and C 9 .
2.2 If x € A fl B then #(&) G #(A) and ^(x) G &(B) and consequently we
see that $(A D 5 ) < ^(^) n #(-B). Now 1?"1 is an automorphism and so
we also have that ^(XnY) < &-l{X) n ^ " ^ F ) . Now put X =
1
and Y = &{B) to get tf" [^(A) n ^(B)] < ^1 n B, whence
n 5 ) . Hence we have the equality #(A n B) = &{A) n
2.8 Let ^>x G InnG be the automorphism <px : g 1-* rc"" 1 ^. Let 1^ G AutG;
we have to show that tf-"1^^ G InnG. Now
But if ^~ x (x) = t/ then ^(z"1) = y"1 and we have
, Book 5 Groups
so ^(pxti G InnG. To show that G/Z{G) ~ Inn<3 define ip : G
Inn (7 by tp(x) = <px-i : g *-+ xgx~l. Then
xygy~lx~l
Clearly, ip is surjective, and
a: G KerV> = • (Vgr 6 G) xgx~x = g = > z G
Thus Ker^ = Z(G) and InnG = Imip - G/Kerip = G/Z{G).
Clearly, 62 <G Z(i? 8 ) since a'lb2a = ( a ' ^ a ) 2 = (6" 1 ) 2 = 62. Now
we have D8 = {I,bib21b~"11a,ab, ab2}ab~1} and a, 6 ^ #(I?8) since oth-
erwise Dg would be abelian. Hence b~x £ Z(D 8 ). Since 62 G Z(DS)
and a ^ Z(Dg) we have a62 ^ Z{Dg). Also ab,ab~1 commute with
a if and only if 6 commutes with a, so ab,ab~x £ Z(DS). Hence
Z(D8) = {b2) ~ C 3 .
= (a, 6
Similarly, Z(<?8) = <z2) a C2 and Q8/Z{Q8) z± C2 x C 2 .
^ o t n questions have a negative answer. The same group, namely the
dihedral group of order 8, serves to provide counter-examples.
Let D8 = (a,6 | a2 = 1, 64 = 1, {ab)2 = 1). If D8 were the direct
product of two non-trivial subgroups, one must be of order 4 and the
other of order 2. But a normal subgroup of order 2 is central, so Dg =
A x B where \A\ = 4 and 5 = Z(DS) = (&2). Now b £ A since
A fl B = {1}, so Ab G A J / A ~ C 2 . Hence (46) 2 = A62 = A, showing
that 62 G A. Thus 5 C A, which contradicts i n 5 = {1}. Thus D8 is
indecomposable.
However, there is a subgroup and a quotient group of Dg each of which
is isomorphic to C2 x C 2 ; for the subgroup, take (a, b2) (or (a6,62)), and
note that D8/Z(DS) ~ C2 x C 2 since Z(D 8 ) = (^2)-
Since G/Z(G) is cyclic, every element is a power of a single element, say
aZ(G). Thus, given g G G, we can write # = a n 2 for some z G
and n G Z. If gi,g2 € G then, with an obvious notation, we have
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2.1 Let \G\ = pa. The class equation gives
where g'x G Cx for \CX\ > 1. Now p divides both \G\ and \G : M{g'x)l so
p divides |Z(G)|. Hence Z(G) is non-trivial.
Now suppose that \G\ = p2. Since Z[G) is non-trivial we have
\Z{G)\ =pov \Z{G)\ = p2. Now if |Z(G)| = p then |G/Z(G)| = p so is
cyclic. Let aZ(G) be a generator of G/Z(G). Then two arbitrary ele-
ments of G are a m x and any where x,y G Z(G). But amxany = anyamx
since re, t/ £ Z(G). Hence G is abelian, and so \G\ = \Z(G)\ — p which is
a contradiction. It follows that we must have |#(G)| = p2 — \G\ whence
G is abelian.
The groups of order 9 are C3 x C 3 and C 9 .
2.2 If x € A fl B then #(&) G #(A) and ^(x) G &(B) and consequently we
see that $(A D 5 ) < ^(^) n #(-B). Now 1?"1 is an automorphism and so
we also have that ^(XnY) < &-l{X) n ^ " ^ F ) . Now put X =
1
and Y = &{B) to get tf" [^(A) n ^(B)] < ^1 n B, whence
n 5 ) . Hence we have the equality #(A n B) = &{A) n
2.8 Let ^>x G InnG be the automorphism <px : g 1-* rc"" 1 ^. Let 1^ G AutG;
we have to show that tf-"1^^ G InnG. Now
But if ^~ x (x) = t/ then ^(z"1) = y"1 and we have
, Book 5 Groups
so ^(pxti G InnG. To show that G/Z{G) ~ Inn<3 define ip : G
Inn (7 by tp(x) = <px-i : g *-+ xgx~l. Then
xygy~lx~l
Clearly, ip is surjective, and
a: G KerV> = • (Vgr 6 G) xgx~x = g = > z G
Thus Ker^ = Z(G) and InnG = Imip - G/Kerip = G/Z{G).
Clearly, 62 <G Z(i? 8 ) since a'lb2a = ( a ' ^ a ) 2 = (6" 1 ) 2 = 62. Now
we have D8 = {I,bib21b~"11a,ab, ab2}ab~1} and a, 6 ^ #(I?8) since oth-
erwise Dg would be abelian. Hence b~x £ Z(D 8 ). Since 62 G Z(DS)
and a ^ Z(Dg) we have a62 ^ Z{Dg). Also ab,ab~1 commute with
a if and only if 6 commutes with a, so ab,ab~x £ Z(DS). Hence
Z(D8) = {b2) ~ C 3 .
= (a, 6
Similarly, Z(<?8) = <z2) a C2 and Q8/Z{Q8) z± C2 x C 2 .
^ o t n questions have a negative answer. The same group, namely the
dihedral group of order 8, serves to provide counter-examples.
Let D8 = (a,6 | a2 = 1, 64 = 1, {ab)2 = 1). If D8 were the direct
product of two non-trivial subgroups, one must be of order 4 and the
other of order 2. But a normal subgroup of order 2 is central, so Dg =
A x B where \A\ = 4 and 5 = Z(DS) = (&2). Now b £ A since
A fl B = {1}, so Ab G A J / A ~ C 2 . Hence (46) 2 = A62 = A, showing
that 62 G A. Thus 5 C A, which contradicts i n 5 = {1}. Thus D8 is
indecomposable.
However, there is a subgroup and a quotient group of Dg each of which
is isomorphic to C2 x C 2 ; for the subgroup, take (a, b2) (or (a6,62)), and
note that D8/Z(DS) ~ C2 x C 2 since Z(D 8 ) = (^2)-
Since G/Z(G) is cyclic, every element is a power of a single element, say
aZ(G). Thus, given g G G, we can write # = a n 2 for some z G
and n G Z. If gi,g2 € G then, with an obvious notation, we have
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