Question

Let $\mathit{G}$ be a non-abelian group of order 10.

1. Prove that$\mathit{G}$contains an element of order 5.

Short answer

It is proved that, $G$contains an element of order 5.

Step-by-step solution

Determine G contains an element of order 5

Consider the given function,$G$is a non-abelian group of order 10.

Now let’s assume that,$G$does not have an element of order 5.

As$\left|G\right|=10=2\times 5$ .

By Lagrange’s theorem, the order of any non-identity element of$G$ must be$2$ or$5$ .

Thus, it is clear that each non-trivial element of $G$has order 2.

That means, for any$g\in G$ with $g\ne e$,

$g=g^{-1}$

Further simplification 

Now claim that $a,b\in G/\left\{e\right\}$,

$ab=ba$

Thus, every element $G$of has order 2,

$ab={\left(ab\right)}^{-1}=b^{-1}{a}^{-1}=ba$

Here the last equality holds as $a^2=b^2=e$. Hence complete the proof of claim. But $G$is non-abelian thus it is not possible.

Therefore, $G$contains an element of order 5.

Hence proved.