# Why am I allowed bring the exponent inside the bracket for the image of a homomorphism…

I am looking at a proof regarding if $\alpha : G \to H$ is a homomorphism, the order of $\alpha(g)$ divides the order of $g$, for $g \in G$.

So I let $|g| = n$.

Then

$\alpha(g)^n = \alpha(g^n) = \alpha(1_G) = 1_H$

So $n$ is a multiple of $|\alpha(g)|$.

That makes sense but what I am not sure about is how we are allowed to bring the exponent $n$ inside the bracket...$\alpha(g)^n = \alpha(g^n)$

We are saying performing some binary operation on the image of $g$ $n$ times is the same as performing the binary operation on $g$ $n$ times and then taking the image of it. Why are we allowed to say this?

Edit: Would someone mind showing me how it would work for $g^3$ or $g^4$ $\ldots$ as it is exponents greater than 2 that are giving me problems.

-
What does it mean: 'homomorphism'? It means that $\alpha(gg')=\alpha(g)\alpha(g')$ for any $g,g'$.. So, apply for $g'=g$, and apply it more.. –  Berci Oct 22 '12 at 23:03
Induction on $n$. –  Mustafa Gokhan Benli Oct 22 '12 at 23:05

By definition of a homomorphism for $a,b\in G$,$\alpha(ab)=\alpha(a)\alpha(b)$. So if we look at $\alpha(a^2)=\alpha(a)\alpha(a)$. Then we can say $\alpha(a^n)=\alpha(a^{n-1}a)=\alpha(a^{n-1})\alpha(a)$

-
But when you say: $\alpha(a^n)=\alpha(a^{n-1}a)=\alpha(a^{n-1})\alpha(a)$ --- But aren't you 'now in $H$' in the right hand side of the equation? So you can't re-apply the same process repeatedly for the remaining $(n-1)$ $a's$..or can you? –  Jim_CS Oct 22 '12 at 23:17
@Jim_CS: The right-hand side is a product in $H$, yes -- but one of the factors of the product is $\alpha$ applied to $a^{n-1}$, and $a^{n-1}$ is still computed in $G$. So to progress you now look at $\alpha(a^{n-1})$ alone, temporarily ignoring that its result is eventually going to be multiplied by $\alpha(a)$ in $H$. –  Henning Makholm Oct 22 '12 at 23:34
@Jim_CS: You can prove that $\alpha(a^n) = \alpha(a)^n$ for all $n \in \mathbb{N}$ by induction. That is what axblount's answer is implicitly doing. –  Michael Joyce Oct 22 '12 at 23:36

$\phi (g^2)=\phi (g) \phi(g)=\phi(g)^2$

-
I can see how it works for $g^2$ from the definition of a homomorphism but I can't see clearly how it works for $g^k$, $k>2$. Would you mind editing in an example for $g^3$? –  Jim_CS Oct 22 '12 at 23:21
@Jim_CS: $\phi(g^3)=\phi(ggg)=\phi(g)\phi(gg)=\phi(g)\phi(g)\phi(g)=\phi(g)^3$ –  Henning Makholm Oct 22 '12 at 23:27
Once you see why this works just use induction as mentioned above. –  BobaFret Oct 22 '12 at 23:48

"We are saying performing some binary operation on the image of $g$ $n$ times is the same as performing the binary operation on $g$ $n$ times and then taking the image of it. Why are we allowed to say this?"

This is exactly (a special case of) what it means to be a homomorphism. A function $\phi: G\rightarrow H$ is a homomorphism if you get the same answer by multiplying together any number of elements in $G$ and then taking their image under $\phi$ as if you take their images under $\phi$ first and then multiply the images (in the same order) in $H$ afterwards.

In the standard definition of homomorphism, we only bother stating that $\phi(gh) = \phi(g)\phi(h)$ for all $g,h\in G$, because it follows by induction from this that what I said above holds (if you don't see that, try writing down a proof).

-