Artin Proposition 2.5.3 This is a rather obvious fact that I know how to prove, but want to be sure that I have proper proof etiquette. The statement is:

If $\varphi: G \to G'$ is a group homomorphism and $a_1, \ldots, a_k \in G$, then $\varphi(a_1, \ldots, a_k) = \varphi(a_1) \cdots \varphi(a_k)$.

This can be proved fairly easily by induction on $k$.

We induct on $k$. The $k = 1$ is clear, and the $k =2$ case follows from the definition of a group homomorphism. Suppose $\varphi(a_1 \cdots a_k) = \varphi(a_1) \cdots \varphi(a_k)$ for some $k \geq 1$. We then have
\begin{align*}
\varphi\left(a_1 \cdots a_k  a_{k+1}\right) = \varphi\left((a_1 \cdots a_k) a_{k+1}\right) = \varphi(a_1 \cdots a_k) \varphi(a_{k+1}) = \varphi(a_1) \cdots \varphi(a_k) \varphi(a_{k+1}),
\end{align*}
which closes the induction.

My question is: should I suppose the result holds for $k \geq 2$? Otherwise, when I write $\varphi(a_1 \cdots a_k a_{k+1})$, it seems that I'm referring to three distinct elements $a_1, a_k, a_{k+1}$ when in fact, if $k = 1$, we have $a_1 = a_k$. I could use product notation to make this transition somewhat more seamless, but then there's some added confusion, as that in s some sense "assumes" commutativity.
 A: How does strong induction work? You want to prove that $P(k)$ is true for all $k \geq 1$. This consists in two steps:

*

*showing that $P(1)$ is true.

*showing that for all $k \geq 1$, $(P(1) \wedge \cdots \wedge P(k)) \implies P(k+1)$.

The predicate $P(k)$ here is "for all $a_1,\ldots, a_k \in G$, $\varphi(a_1\cdots a_k) = \varphi(a_1)\cdots \varphi(a_k)$". So for step $1$, we must take $a_1 \in G$ and show that $\varphi(a_1)$ (the left side) equals $\varphi(a_1)$ (the right side). Nothing to do.
For step $2$, fix $k \geq 1$, assume that $P(1)$ through $P(k)$ are all true, and show that $P(k+1)$ is true. As you observe, there are two cases to look at. If $k=1$, we only assume that $P(1)$ is true, and have to prove that $P(2)$ is true. Of course, $P(1)$ alone is completely useless, but $P(2)$ being true is saying that $\varphi$ is a homomorphism, which was our assumption. This concludes this case. If $k \geq 2$, take $a_1,\ldots, a_{k+1} \in G$, and proceed as $$\varphi(a_1\ldots a_{k+1}) = \varphi((a_1\cdots a_k)a_{k+1}) \stackrel{P(2)}{=} \varphi(a_1\ldots a_k)\varphi(a_{k+1}) \stackrel{P(k)}{=} \varphi(a_1)\cdots \varphi(a_{k+1}).$$
