Question

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

Short answer

We proved that the groupG of order 8 contains an element of order 2.

Step-by-step solution

To determine the 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,

The 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:

Case 1: o(x)=8

Take element $y=x^4$.

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

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

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

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, there is nothing to prove.

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