Question

Prove that $\mathit{G}$ contains five elements of order 2.

Short answer

It is proved that, $G$contains five elements of order 2.

Step-by-step solution

Determine G contains five elements of order 2

As part (a) that $G$has one element of order 5, say $a\in G$, then $H:=\{e,a,a^2,a^3,a^4\}$

It is a subgroup of $G$. As $\left|G\right|=10$, here is an element in $G/H$, say $b$. And it follows that, $H\cap bH=\theta$.

Also, $|H\cup bH|=\left|G\right|$.

Implies that,$H\cup bH=G$ . Now claim that for every$0\le i\le 4,b{a}^i$ has order 2.

Further simplification

Let’s consider that, $i\in \{0,1,2,3,4\}$and suppose the element $b{a}^i$. Note that ${\left(b{a}^i\right)}^2\ne b{a}^j$, here $j=0,1,2,3,4$. If for some $j=0,1,2,3,4$${\left(b{a}^i\right)}^2=b{a}^j$then it would be $b{a}^{i}b{a}^i=b{a}^j\Rightarrow b=a^{j-2i}\in H$

Contradicting the fact that $b\in G/H$. Thus ${\left(b{a}^i\right)}^2\in H$for each $i=0,1,2,3,4$.

Also, by observing that ${\left(b{a}^i\right)}^2\ne b{a}^j$, for any $j=0,1,2,3,4$. As it implies that the order of ${\left(b{a}^i\right)}^2$is 5.

Consequently, the order of $b{a}^i$is 10. The $G$ would be cyclic group generated by role="math" localid="1654354271953" $b{a}^i$, this is not possible because $G$ is non-abelian. Thus ${\left(b{a}^i\right)}^2$has to be $e$implies that the order of ${\left(b{a}^i\right)}^2$is 2 for $i=0,1,2,3,4$.

The number of such elements is 5. Hence proved.