Problem
Let \(G\) and \(H\) be groups. Let \(\varphi:G\rightarrow H\) be a homomorphism.
- Prove that \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}^+\).
- Do the same for \(n = -1\) and deduce that \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}\).
Solution
We first prove that \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}^+\).
For \(n = 1\), $$ \varphi(x^n) = \varphi(x^1) = \varphi(x) = \varphi(x)^1 = \varphi(x)^n $$
Suppose that \(\varphi(x^k) = \varphi(x)^k\) for some \(k \in \mathbb{Z}^+\). Then, since \(\varphi\) is a homomorphism,
$$ \varphi(x^{k+1}) = \varphi(x^kx) = \varphi(x^k)\varphi(x) = \varphi(x)^k\varphi(x) = \varphi(x)^{k+1} $$
Therefore, by mathematical induction, \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}^+\).
$$\tag*{$\blacksquare$}$$
Since \(\varphi\) is a homomorphism, it follows that
$$ \varphi(1)\varphi(x) = \varphi(x) = \varphi(x)\varphi(1) $$
That is, \(\varphi\) maps the identity in \(G\) to that of \(H\). In other words, \(\varphi(1) = 1\).
Note this also proves that, when \(n = 0\), $$ \varphi(x^n) = \varphi(x^0) = \varphi(1) = 1 = \varphi(x)^0 = \varphi(x)^n $$
Now consider the case where \(n = -1\). Again, since \(\varphi\) is a homomorphism,
$$ \varphi(x)\varphi(x^{-1}) = \varphi(xx^{-1}) = \varphi(1) = 1. $$
Therefore, \(\varphi(x^{-1})\) is the inverse of \(\varphi(x)\). That is, \(\varphi(x^{-1}) = \varphi(x)^{-1}\).
Finally, consider \(n < 0\). Note we already showed that \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}^+\). It follows that, for \(n < 0\), $$ \varphi(x^n) = \varphi(x^{-|n|}) = \varphi((x^{-1})^{|n|}) = \varphi(x^{-1})^{|n|} = (\varphi(x)^{-1})^{|n|} = \varphi(x)^{-|n|} = \varphi(x)^n $$
Hence, \(\varphi(x^n) = \varphi(x)^n\) for all \(n \in \mathbb{Z}\).
$$\tag*{$\blacksquare$}$$