# There exists an element of order $5$ in $G$

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?

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 $$|\ker\varphi'|$$).
Note that $$\mathbb{Z}/10 \cong G/K$$ for some $$K$$ a normal subgroup of $$G$$. In particular, $$|G| = 10|K|$$, so $$|G|$$ is a multiple of $$5$$. Therefore by Cauchy's theorem, there exists an element of $$G$$ of order $$5$$.
I think you are overcomplicating this. By First Isomorphism Theorem, $$G/N\equiv\Bbb{Z}_{10}$$, where $$N$$ is the kernel of $$\phi$$. However, this means the order of $$G/N$$ is $$10$$. Since the order of $$G/N$$ is a factor of the order of $$G$$, this means the order of $$G$$ is a multiple of $$10$$. Therefore, the order of $$G$$ is a multiple of $$5$$, since $$5$$ is a multiple of $$10$$. Thus, by Cauchy's theorem, $$G$$ contains an element of order $$5$$ because $$5$$ is prime.