$g_{1},g_{2} \in G$, then if $g_{1}g_{2}$ has finite order then we have to prove that $g_{2}g_{1}$ also has finite order too.

Question

Let $G$ be a group and $g_{1},g_{2} \in G$, then if $g_{1}g_{2}$ has finite order then we have to prove that $g_{2}g_{1}$ also has finite order too.

I have proved like this-

If $g_{1}g_{2}$ has finite order that means there exists positive integer $n$ such that $(g_{1}g_{2})^n = e$, now we rewrite the equation in a clever way.

$(g_{1}g_{2})^n = e$

$(g_{1}g_{2})(g_{1}g_{2})(g_{1}g_{2})...(g_{1}g_{2})= e$

$g_{1}(g_{2}g_{1})^{n-1} g_{2} = e$

$(g_{2}g_{1})^{n-1} = g_{1}^{-1}g_{2}^{-1}$

Multiplying $g_{2}g_{1}$ on both sides from the left hand side, we get -

$(g_{2}g_{1})^n = g_{2}g_{1}g_{1}^{-1}g_{2}^{-1} = e$

Implying that the element $g_{2}g_{1}$ has finite order too.

Now I am thinking of another approach, that is through homomorphism -

That is can we find a homomorphism between the two -

Say $\phi$, that is $\phi(g_{1}g_{2}) = g_{2}g_{1}$, which would mean that $\phi(g_{1}) \phi(g_{2}) = g_{2} g_{1}$.

Any candidate for $\phi$, as if I can find a homomorphism then the order of both the elements will be same.

For finding such a homomorphism, try conjugation by either $g_1$ or $g_2$. — Tobias Kildetoft, Oct 10 '18 at 08:08
Like conjugation to which element?, like $g_{1} (g_{1}g_{2}) g_{1}^{-1}$? — BAYMAX, Oct 10 '18 at 08:13
If we take $\phi(g) = xgx^{-1}$, for some $x \in G$ then for $g = g_{1}g_{2}$, and $x = g_{1}$; $\phi(g_{1}g_{2}) = g_{2}g_{1}g_{2}g_{2}^{-1} = g_{2}g_{1}$, is this correct? — BAYMAX, Oct 10 '18 at 08:24
So correcting , If we take $\phi(g) = xgx^{-1}$, for some $x = g_{2}$ then for $g = g_{1}g_{2}$, $\phi(g_{1}g_{2}) = g_{2}g_{1}g_{2}g_{2}^{-1} = g_{2}g_{1}$, is this correct? — BAYMAX, Oct 10 '18 at 08:26
Yes, precisely. And this $\phi$ is even an automorphism, so the elements have the same order. — Tobias Kildetoft, Oct 10 '18 at 08:27
Oh great!! would not a homomorphism would have guranteed the same order of the elements? — BAYMAX, Oct 10 '18 at 08:28

$g_{1},g_{2} \in G$, then if $g_{1}g_{2}$ has finite order then we have to prove that $g_{2}g_{1}$ also has finite order too.

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