Let $G\cong H$. If $\exists g\in G$ s.t. $|g|=n$, then $\exists h\in H$ with $|h|=n$, and if each element of $G$ is finite, then each in $H$ is too.

Question

If $G, H$ are groups and $G$ is isomorphic to $H$, prove the following:

• If $G$ has an element of order $n$, then $H$ has an element of order $n$, and

• If each element of $G$ has finite order, then each element of $H$ has finite order.

For both, isn't there a theorem that states the exact thing? If this is straight definition, what am I suppose to prove?

Answer 1

Every somewhat experienced mathematician in every situation arising in practice will take 'being isomorphic' as meaning 'having all their group theoretic properties to be the same' and so yes, with this definition of isomorphism there is nothing to prove.

HOWEVER: when you are a student, doing a course, there is a different definition of 'isomorphism' that you are using, namely bijective homomorphism where a homomorphism is a map satisfying $f(ab) = f(a)f(b)$.

So what the person giving you this exercise is asking you to do is showing (a few special cases of the fact) that the statement 'there is an isomorphism (in the bijective homomorphism definition) between $G$ and $H$' is indeed equivalent to the more useful statement '$G$ and $H$ are the same group'. You have to go through the trouble of proving this yourself once before you are 'morally' allowed to use the '$G$ and $H$ are the same group' definition of being isomorphic for the rest of your life.

Answer 2

Hint: Show by induction on positive integers $n$ that for an isomorphism $\varphi: G\to H$, we have $\varphi (g^n)=(\varphi(g))^n$ and that for negative integers $m$, $-m$ is positive, so $\varphi (g^\ell)=(\varphi(g))^\ell$ for all $\ell\in\Bbb Z$.

Hint 2: Show that $\varphi (e_G)=e_H$.

Hint 3: (This is more of a suggestion really.) For the second statement, try proving the contrapositive, that if there exists an element of $H$ that has infinite order, then there exists an element of $G$ that has infinite order.