Let $G$ be a finite group and assume $\varphi: \, G \longrightarrow \mathbb{Z}_{10}$ is a group epimorphism. I want to show that there exist an $a \in G$ s.t. $\text{ord}(a)=5$.
At first, the cyclic group:
$$
\langle2\rangle=\{0,2,4,6,8\}
$$
is a subgroup of $\mathbb{Z_{10}}$ of order $5$.
The epimorphism $\varphi$ induces anοther epimorphism $\varphi ': \langle a\rangle \longrightarrow \langle 2 \rangle$, where $a \in G$ s.t. $\varphi(a)=2$. By the fundamental homomorphism theorem:
$$
\langle 2 \rangle \cong \langle a\rangle/\text{ker}\varphi '\iff |\langle a\rangle|=5 \cdot |\text{ker}\varphi '|
$$
Is there a way to prove that $|\text{ker}\varphi '|=1$ for some $a \in G$? In other words, how can it be shown that there exists an $a \in G$ which makes $\varphi '$ an isomorphism?
Best Answer
There's a far easier option: since $\varphi$ is epic, it's surjective, so there is some $g \in G$ such that $\varphi(g) = 2$, and the order of $2$ in $\mathbb{Z}_{10}$ is $5$, so the order of $g$ in $G$ is a multiple of $5$ (there is some $n$ such that $g^n = 1_G$ since $G$ is finite, and if $n = 5a+b$, then $0 = \varphi(1_G) = \varphi(g^{5a+b}) = \varphi(g)^{5a+b} = 2^{5a+b} = 2^b$, so $b$, hence $n$, is a multiple of $5$).
So $|g| = 5k$ for some $k$, and $|g^k| = 5$.
You can do a similar thing with your approach: once you know that $\varphi'$ is an epimorphism, you know that $5$ divides the order of $a$, so essentially the same argument tells you that the order of some power of $a$ is $5$ (and just to answer your final question: yes, just raise your original $a$ to the power of $|\ker\varphi'|$).