About the proof of Cayley’s Theorem

Question

I'm reading a proof of Cayley's theorem. In the following two snippets I want to know why it's needed(although it's apparently true) that $\pi_a\circ\pi_b=\pi_{ab}$ to prove that $G^*$ is closed under composition?

I meant, why not just said: Since $a,b\in G, ab\in G$ so by the construction of $\pi_a:G\to G$ clearly $\pi_{ab}\in G^*$.

At the end of the proof the equality is used, so I wonder whether the author just want to condense two steps into one? (I skip the part of $G^*$ is closed on its inverse)

Answer 1

You're right that it is clear that $\pi_{ab} \in G^*$. But we want $G^*$ to be closed under the group operation of $G^*$, in which the product of $\pi_a$ and $\pi_b$ is defined to be $\pi_a \circ \pi_b$, not $\pi_{ab}$.

Once we prove that these two are equal - that $\pi_a \circ \pi_b = \pi_{ab}$ - then we know that $G^*$ is closed under $\circ$, precisely because we know that $\pi_{ab} \in G^*$.

Answer 2

To prove that $G^*$ is a subgroup of the group of permutations on the set $G$ one should prove that for every $a,b \in G$ there is a $c \in G$ such that $$\pi_a\circ\pi_b=\pi_{c}$$

So the proof needs to provide a way to choose a $c\in G$ such that indeed $\pi_a\circ\pi_b=\pi_c$ holds. The element $c=ab$ is a natural choice.

The apparent complication with this subgroup stuff is due to the fact that the authors wanted to have $G^*$ as the image of the homomorphism, probably because they are interested to show that $\pi$ is an isomorphism.

A different approach could have been to see $\pi$ as an injective homomorphism from $G$ into the group of permutations on $G$ and then simply use that fact that the image of this homomorphism (namely $G^*$) is isomorphic to the starting group ($G$). In this case one use the fact that the image of an homomorphism is a group to prove that $G^*$ is a group.

Hope this helps.