Prove that every finite group having more than two elements has a nontrivial Automorphism.
It is from Topics in Algebra by Herstein. I am not able to solve.
Asked
Active
Viewed 2,475 times
2
-
try $x\mapsto a^{-1}xa$. – May 14 '13 at 14:29
-
Do you see how to solve this when some element has order $>2$? – Ted Shifrin May 14 '13 at 14:30
-
1@CutieKrait, great unless $G$ is abelian ... :) – Ted Shifrin May 14 '13 at 14:32
-
@TedShifrin: then try $x\mapsto x^p$ for an appropriate $p$. – May 14 '13 at 14:42
-
@CutieKrait, great unless every element of $G$ has order $2$ ... :) – user1729 May 14 '13 at 17:31
-
@user1729: then solve the problem for $\Bbb Z_2^n$ for $n>1$. – May 14 '13 at 17:44
-
@CutieKrait: indeed. – user1729 May 14 '13 at 19:39
1 Answers
5
Suppose that $G$ is abelian with exponent greater than 2 (i.e. not every element has order 2). Then mapping $g\in G$ to $g^{-1}$ is an automorphism. It's clearly bijective, and since $G$ is abelian it is indeed a homomorphism.
If $G$ is abelian with exponent 2, then it is a vector space over $\mathbb Z/2$. Choose a basis for this vector space, and pick your favourite two basis vectors to interchange. This is a nontrivial automorphism.
If $G$ is not abelian, then the center of $G$, $Z(G)\neq G$. Now, let $G$ act on itself via conjugation. These are all automorphisms. Convince yourself that if $Z(G)\neq G$, there is a nontrivial one.
Dhruv Ranganathan
- 1,303