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In proving the set of automorphisms of a group $G$ is a group, I need to verify the existence of an identity element. I can manage to show that the identity map is an element of this group, but demonstrating that it performs the operation of an identity (which should be the easiest part) is confusing.

First, we take the map $e: G \to G$ defined by $x \mapsto x$. This map is surjective since for any $y \in G$, $\exists x = y$ such that $e(x) = y$. This map is injective since $e(x) = e(y)$ implies $x = y$ for any $x, y \in G$. Thus, $e$ is bijective. Since for any $x, y \in G$, we have \begin{align*} e(xy) = xy = e(x)e(y), \end{align*} $e$ possesses the homomorphism property. Thus, $e$ is an isomomorphism from $G$ to itself, and is hence an element of ${\rm Aut } G$.

The role that this map $e$ should play is that for any $f \in {\rm Aut }G$, we have \begin{align*} e \circ f = f \circ e = f. \end{align*} For the map $f: G \to G$, $x \mapsto y$, where $f \in {\rm Aut } G$, we have \begin{align*} (e \circ f)(x) &= e(f(x)) = e(y) = y \\ (f \circ e)(x) & = f(e(x)) = f(x) = y. \end{align*} Hence, $e \circ f = f \circ e$ since $x \in G$ was arbitrary.

Is this a valid proof? Have I missed anything?

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  • $\begingroup$ Did you rename $i$ to $e$ on the fly? $\endgroup$ – Christoph Feb 14 at 8:02
  • $\begingroup$ In the second last equation it should be $f(x)$ not $e(x)$. $\endgroup$ – PrudiiArca Feb 14 at 8:08
  • $\begingroup$ @PrudiiArca It's $e(y)$, but I agree the notation is confusing. $\endgroup$ – Christoph Feb 14 at 8:09
  • $\begingroup$ I accidently used $i$ and $e$ interchangeably. Sorry about that. I'll use more consistent notation. $\endgroup$ – John P. Feb 14 at 8:11
  • $\begingroup$ I apologize. Why is it $f(x)$? I must be missing something. $\endgroup$ – John P. Feb 14 at 8:12
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Your proof is fine. I'd just suggest two modifications:

  1. Instead of calling the identity map on $G$ by names like $i$ or $e$, just stick to the usual convention of $\operatorname{id}_G$.
  2. When you consider an arbitrary automorphism $f\colon G\to G$, don't define it as "$x\mapsto y$". This "$y$" depends on the choice of $x$, and that gets lost in your notation. Just keep referring to it as $f(x)$.

Hence, your equations become:

\begin{align*} (\operatorname{id_G}\circ f)(x) &= \operatorname{id_G}(f(x)) = f(x), \\ (f\circ \operatorname{id_G})(x) &= f(\operatorname{id_G}(x)) = f(x). \\ \end{align*}

Furthermore, in the conclusion you need $$ \operatorname{id_G}\circ f = f\circ\operatorname{id_G} = f. $$ You missed the last part.

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