Solution verification: $G$ and $G/H$ contain elements of same order

Question

I just took my abstract algebra midterm, and was wondering if someone could confirm my solution to the following problem.

Problem: Let $G$ be a finite group and let $H$ be a normal subgroup of $G$. Let $n \geq 1$ be an integer and suppose there exists an element of order $n$ in $G/H$. Prove that there exists an element of order $n$ in $G$.

My Solution: Since $H$ is a normal subgroup, there exists a surjective homomorphism $\varphi: G \to G$ such that $\ker(\varphi) = H$. By the First Isomorphism Theorem,

$$ G/\ker(\varphi) \cong \varphi(G). $$

Since $\ker(\varphi) = H$ and $\varphi(G) = (G)$ (since $\varphi$ is surjective), this becomes

$$ G/H \cong G $$

which implies that

$$ |G/H| = |G|. $$

Since $G/H$ and $G$ are of the same order, if $G/H$ contains an element of order $n$, $G$ must as well.

Answer 1

Hint: If $a\in G/H$, there exists $g\in G$ such that $g\mapsto a$ under the canonical map $G\to G/H$. What can you say about the orders of $a$ and $g$? If $g^m=1$, what about $a^m$?

Answer 2

If $(gH)^m=H$, so $gH$ has order $m$ in G/H, then $g^m\in H$ (why?). If $g^m=1$, then we are done, otherwise $g^m\in H$ has some order, $n$ say. So $g^{mn}=1$. Consider $g^n$...

This "breaks" for infinite groups, as $g$ might have infinite order in $G$. For example, consider $G=\mathbb{Z}$ and $H$ any non-trivial subgroup.

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Also, I should say that your concluding sentences incorrect, so your attempt is unfortunately wrong. A counter-example to the last line is the groups of order 4: The Klein 4-group has order 4, as does the cyclic group of order 4, but the Klein 4-group does not contain en element of order 4 while the cyclic group does.