Question

Prove that no subgroup of order 2 in ${\mathit{S}}_{\mathbf{n}}\mathbf{(}\mathbf{n}\mathbf{\ge }\mathbf{3}\mathbf{)}$ is normal.

Short answer

It is proved that, there is no subgroup of order 2 in$S_n$ with$n\ge 3$ is normal.

Step-by-step solution

Determine that no subgroup of order 2 in Sn(n≥3) is normal

Let’s consider that $\sigma \in S_n$with $n>2$is the center of $S_n$and $\sigma \ne e$.

Then, there exists $i\in \mathbf{\{}1,....,n\mathbf{\}}$so that $\sigma \left(i\right)=j\ne i$for $j\in \{1,\mathrm{....},n\}$.

As $n>2$then, there exists $k\in \{1,\mathrm{...},n\}$so that, $i\ne j\ne k$.

Evaluate $\sigma \circ \left(ik\right)$at $i$as:

$\sigma \circ \left(ik\right)\left(i\right)=\sigma \left(k\right)$

But as $\mathit{\sigma }$is the center of $S_n$, then we have:

$\begin{array}{c}\sigma \circ \left(ik\right)\left(i\right)=\left(ik\right)\circ \sigma \left(i\right)\\ =\left(ik\right)\left(j\right)\\ =j\end{array}$

Therefore,$\sigma \left(i\right)=\sigma \left(k\right)$ implies $\sigma \left(k\right)=j$.

This is not possible as $i\ne k$and $\sigma \in S_n$.

It is verifying that, the center of $S_n$is$\left\{\mathrm{e}\right\}$ .

Determine further simplification

Now, if there is normal subgroup of $K$of order 2 in $S_n$with $n\ge 3$, it implies that $K$contains the center of role="math" localid="1654607805007" $S_n$.

As K is the normal subgroup,$\sigma K{\sigma }^{-1}\subset K$for all $\sigma \in S_n$.

Consequently,$\sigma K{\sigma }^{-1}=e$or $\sigma K{\sigma }^{-1}=K$.

If $\sigma K{\sigma }^{-1}=e$then $k=e$, and this is not possible as the order of $k$is 2.

Therefore, $\sigma k{\sigma }^{-1}=\sigma$ implies $\sigma k=k\sigma$.

Any $\sigma \in S_n$implies that $K\subset Z\left(S_n\right)$.

But $K$has order 1 which shows that $Z\left(S_n\right)=\left\{e\right\}$.

Therefore, there is no normal subgroup of order 2 of $S_n$with$n\ge 3$ .