# Let $G$ be a finite group and $\phi:G \to K$ be a surjective homomorphism and $n \in \mathbb{N}.$ If $K$ has an element of order $n,$ so does $G.$

Let $G$ be a finite group$\ ,\phi:G \to K$ be a surjective homomorphism and $n \in \mathbb{N}.$ If $K$ has an element of order $n,$ so does $G.$

May I know if my proof is correct? Thank you.

Proof: Let $w \in K$ such that $o(w)=n.$ Also, $\ \exists g \in G$ such that $w = \phi(g),$whence $w^n = \phi(g^n)=e_K.$

Since $|G| < \infty, o(g^n) \leq |G|.$ Let $m = \min\{n' \in \mathbb{N}:n' \leq |G| \ \wedge g^{nn'} = e_G\}.$

Given any $s \in \mathbb{N}$ such that $g^s=e_G,$ then $\phi(g^s)=w^s=e_K,$whence $n|s.$

Since $(g^n)^{\frac{s}{n}} = e_G$ and $m$ is the minimum element, $\frac{s}{n} \geq m,$ whence $s \geq mn$

Hence, $o(g) =mn$ and $o(g^m)=n.$

• Proof looks good to me. I don't think you need the condition $n' \leq |G|$ on the RHS of the definition of $m$ (certainly the minimum $n'$ satisfying the second condition is automatically bounded by $|G|$). Commented Nov 25, 2013 at 14:51
• @Martin Brandenburg: Ok, I have edited. Commented Nov 25, 2013 at 14:51

The proof is OK. But you might use the well known and generally useful fact that a cyclic group of order $m$ has a cyclic subgroup (and hence elements) of any order $d$ dividing $m$, to reduce this to a one-liner. With $o(w)=n$, $\phi(g)=w$ and $m=o(g)$ one has $e_K=\phi(e_G)=\phi(g^m)=w^m$, so $n$ divides $m$, and the cyclic subgroup $\langle g\rangle$ of $G$, which has order $m$, contains an element of order $n$.