Question

Prove that a group of order 8 must contain an element of order 2.

Short answer

We proved that, the group $G$of order 8 contains an element of order 2.

Step-by-step solution

To determine order of x.

Let $G$be a group and$x\in G$ be any element of $G$.

Let $o\left(G\right)=8$.

We have to show that$G$ must contain an element of order 2.

Since $o\left(G\right)=8$, by Lagrange’s Theorem ,

Order of every element must divide $o\left(G\right)$.

Let $x\in G$such that $x\ne e$.

Then$o\left(x\right)$ is either 1, 2, 4 or 8.

Since $x\ne e$,$o\left(x\right)\ne 1$ .

Now, here we consider two cases:

Case1: o(x)= 8

Take element $y=x^4$

Since $o\left(x\right)=8$, $y\ne e$

Then$y^2={\left(x^4\right)}^2=x^8=e$

$\Rightarrow o\left(y\right)=2$

$y^2={\left(x^4\right)}^2=x^8=e$

Case 2: o(x)= 4

Take element $y=x^2$.

Since $o\left(x\right)=4$, $y\ne e$.

Then $y^2={\left(x^2\right)}^2=x^4=e$

$\Rightarrow o\left(y\right)=2.$

Next, if$o\left(x\right)=2$ already, then nothing to prove.

Hence, the group $G$of order 8 contains an element of order 2.