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I was trying to show that there exist a nontrivial automorphism $\pi$ in $\text{Aut}(G)$, and I am taking the case that $G$ is abelian because if $G$ is nonabelian it is trivial. If there exists an $\exists{x}$ with $|x| \neq 1,2,$ then we have the automorphism $\pi_{-1}(g) = g^{-1}$. My question is why must we look at the case where there exists an $\exists{x}$ with $|x| \neq 1,2,$ and how does that imply the automorphism $\pi_{-1}(g) = g^{-1}$?

Martin Sleziak
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John Ryan
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1 Answers1

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If every $g \in G$ has order $1$ or $2$, then $$g^{-1}=g \forall g \in G$$ which means that $g \to g^{-1}$ is trivial.

Now, if some particular $x$ has order different of 1 or 2, then $$\pi_{-1}(x)=x^{-1} \neq x$$ which shows that $\pi_{-1}$ is not trivial...

Showing that $\pi_{-1}$ is an automorphism is nor related to the order, but it is an easy exercise in any Abelian group. The problem is that it could be (and in groups where each element has order 1,2 it is) the trivial automorphism.

N. S.
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  • So they just do that in order to make sure it isn't a trivial automorphism. – John Ryan Jan 18 '16 at 18:46
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    @JohnRyan Yes, that is what it means to be non-trivial. To make it clearer, le $Id : G \to G$ be $Id(g) =g$. This is called the trivial automorphism. Non trivial means $$\pi_{-1} \neq Id$$. – N. S. Jan 18 '16 at 18:48
  • Also, one more thing. When we take the automorphism of say $\mathbb{Z}/2$, we aren't assuming any placeholder operation such as addition, multiplication, etc.? – John Ryan Jan 18 '16 at 18:50
  • @JohnRyan Yes.. In this case your group is supposed to be $(G, \cdot)$, where $\cdot$ is a binary operation which makes it a group. Automorphism is with respect to this operation.... – N. S. Jan 18 '16 at 18:52
  • Ah, that makes sense. So if we say for example $\text{Aut}(G) \simeq GL_{n}(\mathbb{Z}/2)$, we aren't indicating any binary operation (so for all operations it is isomorphic)? Or I think it is the composition of functions because of $\text{Aut}(G)$? – John Ryan Jan 18 '16 at 18:53
  • I think my question is if the binary operation is not given, is it implied what operation we are using or is there sometimes when we are talking about any operation? – John Ryan Jan 18 '16 at 19:01
  • @JohnRyan A group is a set with a binary operation. So the binary operation is implicitly assumed, even if not explicitly given. Whenever when you see a group you know it is a set $G$ with some binary operation.... – N. S. Jan 18 '16 at 19:04
  • How can I tell what binary operation is being used? Sometimes it is obvious because if we are working with $\text{Aut}(G)$ that is a group under compositions but $\mathbb{Z}/2$ could be under other operations. So what would I do in the case of $\mathbb{Z}/2$? – John Ryan Jan 18 '16 at 19:07