# The choice of generators and the order of $\operatorname{Aut}(G)$.

A group homomorphism $G \to G$ is fixed by its values on a set of generators of $G$.

Consider the order of the automorphism group $\operatorname{Aut}(G)$.

Let $S = \{s_1,s_2,\ldots,s_n\}$ be a minimal set of generators for $G$. Note that an isomorphism $G \to G$ sends a generator $s_i$ to a generator $s_j$ of the same order.

Counting, per order $k$, the amount of generators $a_k$, we see that there is a total of $$T = \prod_k (a_k!)$$ possible isomorphisms. Namely, for $a_k$ generators of order $k$ we have $a_k!$ ways of permuting the the generators of order $k$ and thus $a_k!$ possible isomorphisms, which send generators to generators of the same order. To get the total number of isomorphisms we take the product of all the the possibilities.

I want to show that the number of isomorphisms $T$ is independent of the choice of generators for $G$.

This seems plausible since the order of $\operatorname{Aut}(G)$ is not determined by the choice of generators for $G$.

PS: I raised this question on account of the homework question $$\text{For two isomorphic groups } G \text{ and } G' \text{, prove that the amount of isomorphisms } \\ \ G \to G' \text{ is equal to the order of the group } \operatorname{Aut}(G) .$$

• Note that an automorphism doesn't need to map all generators $s_j$ into $S$, consider e.g. the quaternion group with $S = \{i,j\}$, an automorphism can map for example $i$ to $k = i\cdot j \notin S$. Also, there may be relations among the generators that an automorphism must respect. The homework question can be answered much more easily by establishing a bijection. – Daniel Fischer Mar 23 '14 at 13:12
• In fact, the set of isomorphisms between $G$ and $G'$ is a principal homogeneous space for $\operatorname{Aut}(G)$. – tomasz Mar 23 '14 at 13:16

Fix a group isomorphism $\phi:G\xrightarrow{\simeq}G^\prime$. Then $\phi$ induces a biiection between the set $\text{Iso}(G,G^\prime)$ of isomorphisms $G\xrightarrow{\simeq} G^\prime$ and the set $\text{Aut}(G)$ of automorphisms of $G$ given by: $$\text{Iso}(G,G^\prime)\to\text{Aut}(G):\xi\mapsto\phi^{-1}\circ\xi$$ with inverse $$\text{Aut}(G)\to\text{Iso}(G,G^\prime):\eta\mapsto\phi\circ\eta.$$
• Can we define an identity element $e \in \operatorname{Iso}(G,G')$ such that $\operatorname{Iso}(G,G')$ becomes a group? Aside from the fact a bijection is sufficient in our case, instead of a group isomorfism $\operatorname{Iso}(G,G') \to \operatorname{Aut}(G)$. – Mussé Redi Mar 23 '14 at 17:16
• @MusséRedi You can use this bijection to define a group structure on $\text{Iso}(G, G')$, in which case the identity is $\phi$. This group operation is not canonical, though. – Dustan Levenstein Mar 23 '14 at 17:35
• Define $\xi_1 \cdot \xi_2$ by $\phi \circ [(\phi^{-1} \circ \xi_1)(\phi^{-1} \circ \xi_2)]$. Canonical is a somewhat vague term mathematicians use to say when a defined function/operation/etc. on an object doesn't depend on some arbitrary choice. – Dustan Levenstein Mar 23 '14 at 17:57