Question

Let $\mathit{N}$ be a subgroup of ${\mathit{S}}_{\mathbf{n}}$such that$\mathit{\sigma }\mathit{\tau }\mathbf{=}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$ for all nonidentity elements $\mathit{\sigma }\mathbf{,}\mathit{\tau }\mathbf{\in }\mathit{N}$. Prove that $\mathit{N}\mathbf{=}\left(1\right)$or$\mathit{N}$ is cyclic of order 2.

Short answer

It is proved that, $N=\left(1\right)$orrole="math" localid="1654608604664" $N$ is cyclic of order 2.

Step-by-step solution

Determine that N(1)or N is cyclic of order 2

Let’s consider that $N$ is the subgroup of $S_n$so that for every non-identity elements role="math" localid="1654608717017" $\sigma ,\tau \in N$and $\sigma \tau =\left(1\right)$.

Let’s suppose that $N\ne \left\{\left(1\right)\right\}$.

Then, there is an element $\sigma \in N$so that $\sigma \ne \left(1\right)$.

As for each and every non-identity elements, $\sigma ,\tau \in N$and $\sigma \tau =\left(1\right)$.

In particular,$\sigma =\tau$ , then, ${\sigma }^2=\left(1\right)$.

This implies that, the order of $\sigma$is 2 as $\sigma \ne \left(1\right)$.

Determine further simplification

Now, from the hypothesis,$\tau \in N$ ,$\sigma \tau =\left(1\right)$ .

Or similarly as:

$\begin{array}{c}\tau ={\sigma }^{-1}\\ =\sigma \end{array}$

Here, the last equality holds as the order of $\sigma$is 2.

Consequently,$N=\{\left(1\right),\sigma \}$ .

From this, it is verifying that$N$ is a cyclic group of order 2.

Hence, it is proved that $N=\left(1\right)$or$N$ is cyclic of order 2.