# For $f:G_1 \to G_2$ a group homomorphism, Prove that $|\{(x,y) \in G_1 \times G_2: f(x^n) = y^n\}| \ \equiv \ 0\mod |G_1|$

I am stuck on the following problem which is from Ex-2.26(b) of these notes.

Given $$f: G_1 \to G_2$$ , a group homomorphism, Where $$G_1 , G_2$$ are finite groups and Let $$n$$ be a fixed positive integer. Then prove that: $$|\{(x,y) \in G_1 \times G_2: f(x^n) = y^n\}| \ \equiv 0\mod |G_1|$$

What I've tried is considering some group action like : $$G_1$$ on $$\ \{(x,y) \in G_1 \times G_2: f(x^n) = y^n\} \$$ defined by $$g * (x,y) = (g \ x \ {g}^{-1},\ f(g) \ y \ f({g}^{-1}))$$ and what I've noticed is for $$(x,y)$$, its orbit is in correspondence to the conjugacy class of $$x$$ in $$G_1. \$$And that's a dead end for me.

So I would like to confirm if this is the 'right' action to consider here and if so how do I make further progress with it, If not then what is the proper way to do these kind of problems? Thanks

• Notice that $f(x^n)=f(x)^n$. Also, the set of $y$-values for a given $x$ is closed under conjugation by $f(x)$. Commented Jul 31 at 23:47
• Hints: reduce to the case $f$ is surjective. Do it for $n=1$ first. Commented Aug 1 at 0:48
• In the case I described, $z$ has to belong to $f(G_1)$ right? Let me describe the solution in my case: we can reduce to the case $f(G_1)=G_2$ since other elements contribute no pairs. Now using $f(x^n)=f(x)^n$, for an arbitrary $z\in f(G_1)$, how many pairs have second coordinate $z$? It's $|\ker(f)|$ times the number of elements $g$ with $g^n=z$. Now sum all these up. Commented Aug 1 at 15:13
• Said another way, how many pairs start with $x$? Well, $z$ is completely determined by $x$, so exactly $1$! Commented Aug 1 at 15:16
• Why would we care if it's injective? Given $x$, there is one $z$, specifically the element $z=f(x^n)$. I've given you two ways to count the pairs. Please spend some time thinking about this until you understand it. Commented Aug 1 at 15:27

Summarizing the hints I gave in chat:

Show that if $$H$$ is a subgroup of the finite group $$G$$, and $$Y=\{g\in G\mid g^n\in H\}$$, then $$|H|$$ divides $$|Y|$$.

Show that $$Y$$ is the union of equivalence classes from 2.25. Use 2.25(c) to conclude.

Continuing with $$H\le G$$, let $$Z=\{(h,g)\mid h^n=g^n\}$$. Show that $$|H|$$ divides $$|Z|$$.

Apply the previous result with $$H\times G$$ the larger group, and the diagonal subgroup $$D=\{(h,h)\mid h\in H\}$$ the smaller group. Then $$Z$$ is just the $$Y$$ from above, and $$|D|=|H|$$.

Solve the problem when $$\ker(f)=\{1\}$$.

In this case, let $$G=G_2$$ and $$H=f(G_1)$$. The previous part shows if $$A=\{(f(g_1),g_2)\mid f(g_1)^n=g_2^n\}$$, then $$|H|=|G_1|$$ divides $$|A|$$. This is enough, since $$f$$ is injective.

Solve the problem.

Let $$B=\{(g_1,g_2)\mid f(g_1)^n=g_2^n\}$$. From the previous part, $$|A|$$ is divisible by $$|f(G_1)|$$. Every element of $$A$$ has $$|\ker(f)|$$ preimages in $$B$$. Thus $$B$$ is divisible by $$|A|\cdot|\ker(f)|=|G_1|$$.