Proving $\varphi(x^n)=\varphi(x)^n$ is a homomorphism for all $n\in\mathbb{N}$.

Question

I began my proof using induction however I'm a little stuck on understanding what to for my inductive step. I'm looking for proof validation and tips to improve my proof-writing.

Pf: Let $G$ and $H$ be groups such that $\varphi :G\longrightarrow H$ be a homomorphism. Then we will proceed by mathematical induction to show that $\varphi (x^n)=\varphi (x)^n$ is true for $n=1$. Then,

$$\varphi (x^n)=\varphi (x)^n \iff \varphi (x^1)= \varphi (x)^1 \iff \varphi(x)=\varphi (x)$$

Then the $\varphi(x^n)=\varphi(x)^n$ is true for the base step. Next let assume that our inductive hypothesis is that $\varphi(x^n)=\varphi(x)^n$ is true for $n=k$.

Then we want to show in our inductive step that $\varphi(x^n)=\varphi(x)^n$ is true for $n=k+1$. Then $$\varphi(x^{k+1})=\varphi(x)^{k+1} \Longleftrightarrow \varphi(x^kx^1)=\varphi(x)^k\varphi(x)^1$$

Now we will multiple both sides by $\varphi(x^{-k})$, then,

\begin{align*} \varphi(x^kx^1)\varphi(x^{-k})=\varphi(x)^k\varphi(x)^1\varphi(x^{-k})\Longleftrightarrow \varphi(x^1)=\varphi(x)^1=\varphi(x)=\varphi(x) \end{align*}

Thus $\varphi(x^n)=\varphi(x)^n$.

I like to say that what I did is correct but I'm not sure if I can justify exactly what happened in the inductive step.

thanks in advance :) and thanks to those who helped me edit this!

Answer 1

There is no need to multiply by $\varphi(x^{-k})$. The defining property of homomorphisms is that $\varphi(ab)=\varphi(a)\varphi(b)$ for any $a,b$ in your group. Insert $a=x^k$ and $b=x$, and you're (almost) done.

That being said, I feel that your last $\iff$ skipped a few steps. It's correct, certainly, but if I were a teacher correcting this, I would say that you missed a huge opportunity to demonstrate that you actually can use the homomorphism property in a calculation (which is half the point of this exercise, the other half being practicing induction). You've just started at the beginning and jumped right to the end in a single step.

Finally, a matter of personal taste: wherever possible, I would prefer a long chain of equalities beginning with $\varphi(x^{k+1})$ and ending in $\varphi(x)^{k+1}$ to a long chain of $\iff$'s beginning with what you want to prove and ending with something trivial. If you do it your way, at least have one equality (with a single $=$) on each line instead of writing them out in one long line. Then it's easier to follow, and you don't really need the $\iff$'s.

Answer 2

You have to prove $\varphi(x^{k+1})=\varphi(x)^{k+1}$ with the induction assumption $\varphi(x^{k})=\varphi(x)^{k}$

So you start with $$\phi(x^{k+1})=\phi(x^kx^1)=\cdots(\because \phi \;\text{is a } \cdots)=\cdots(\because \text{by induction})=\phi(x)^{k+1}$$