You goal is to show H is a subgroup.

The multiplication * in H is inherited from the multiplication * in G, so the associativity property of that multiplication is also inherited.

The work is in showing that H is closed under *, that H is closed under *-inverses, and that the *-identity e is in H.

However, once you've shown that H is closed under *, and under *-inverses (and that H isn't empty), then if x in H, then so is x^-1, and so x*(x^-1) = e.

Thus the only work is to show that H is closed under *, and that H is closed under *-inverses.

Now those two things can actually be combined into a single thing, but since you're not comfortable with this stuff yet, maybe it's best if you show them each separately for now.

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So: You need to show two things. 1) H is closed under *. 2) H is closed under *-inverses.

Your work should look like this:

1) Claim H is closed under *.

Proof:

Let a, b in H.

Then ... (your arguments here)...

Therefore a*b in H. That proves that H is closed under *.

2) Claim H is closed under *-invereses.

Proof:

Let a in H.

Then ... (your arguments here)...

Therefore a^-1 in H. That proves that H is closed under *-inverses.

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The way to proceed is virtually forced. At each step, there's pretty much only one thing to say, and saying it will get you to your result.

In both proofs, you begin with something, x, in H. What can you do with that? There's only one thing conceivable to say next, because there's only one thing you know when x in H, and that's what the definition of H tells you.

So when you have x in H, you then basically have no choice but to write on the next line: "Therefore, f(x) is in H'."

OK, so then what? Well, you know nothing about H' except it's a subgroup of G'. But that tells you a lot. It tells you that H' is closed under multiplications and inverses in G'.

Not to give it all away to directly, from there, you'll have only one path to follow - use that f is a homomorphism.

From there, you'll have only one path to follow - use the definition of H again. At that point, you will have reached your desired conclusion and the claim will be proven.

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The idea of the proof is that we can use that " push it into H' ", then "use that H' is a subgroup" then "bring it back to H because f is a homomorphism" so that all those nice properties hold for H as well.