Let $e$ and $f$ be in group $G$. Prove $(ef)^n=e(fe)^{n-1}f$

Question

$G$ is a group. Let $e$ and $f$ be in $G$. Use induction to prove that $(ef)^n=e(fe)^{n-1}f$ for all $n>0$.

For the base case with $n=1$, I proved $(ef)^1=e(fe)^{1-1}f$ which gives $(ef)^1=e(fe)^{0}f$ so $ef=ef$

I am not sure how to proceed for $n+1$. Possibly by splitting up $(ef)^{n+1}$ to be $(ef)^n(ef)$?

Answer 1

You don't need to use induction, just spell it out:

$$ (ef)^{n}=(ef) \cdots (ef) =e(fe) \cdots (fe)f =e(fe)^{n-1}f $$

Edit: since your teacher asked you to do so, you can make explicit the underlying induction:

• For $n=1$, the claim follows from $(fe)^{0}=1$.

• If the claim is true for $n \geqslant 1$, then $(ef)^{n+1}=ef(ef)^{n}=ef(e(fe)^{n-1}f)=e(fe)(fe)^{n-1}f=e(fe)^{n}f$.

Answer 2

For a more explicitly induction based proof, the induction step might go:

$(ef)^{n+1}=(ef)(ef)^n = (ef)e(fe)^{n-1}f = e (fe) (fe)^{n-1}f =e (fe)^n f $

Answer 3

Suppose $(ef)^m=e(fe)^{m-1}f$, and you want to prove that $(ef)^{m+1}=e(fe)^mf$

Let's start from LHS,

\begin{align} (ef)^{m+1}&=(ef)^mef \\ &=e(fe)^{m-1}fef \\ &= e(fe)^{m-1}(fe)f \\ &=e(fe)^mf \end{align}

and you reaches RHS.

Answer 4

There are good answers already, but let me elaborate on the point where exactly is induction needed.

First of all, what does $x^n$ mean?

Common way to define it is recursively:

$$x^1 = x,\ x^{n+1} = x^n x.$$

Unwrapping it we get $$x^n = ((\ldots(((xx)x)x)\ldots )x)x.$$ You might wonder what's going on here, isn't it just $x^n = xx\cdots x$? Well, yes and no, what we are doing here is abusing the notation because of associativity. But, sure, let's say that we can just accept that because of associativity $x^n = xx\cdots x$ is unambiguous and never again bother with it.

But, here you have $(ef)^n = (ef)(ef)\cdots (ef)$ and what you want to change it to is $e(fe)\cdots (fe)f$. If you want to do it for particular $n$, say $n =15$, you can do it just by applying associativity long enough until you get what you want. But, here's the deal, you are not tasked to do it for particular $n$, you are tasked to prove that this property holds for all positive integers. And you do need induction to do that, even if it is present only implicitly.