Element of Group Not Conjugate to Proper Subgroup

Theorem
Let $G$ be a finite group.

Let $H$ be a proper subgroup of $G$.

Then there is at least one element of $G$ not contained in $H$ or in any of its conjugates.

Proof
Let $S \subseteq G$ be defined by:


 * $S := \left\{{g \in G: \exists h \in H, a \in G: g = a h a^{-1}}\right\}$

It is required to show that $S \ne G$.

Let $N_G \left({H}\right)$ be the normalizer of $H$ in $G$.

By Subgroup is Subgroup of Normalizer, $H \le N_G \left({H}\right)$.

Therefore, by definition of index:


 * $\left[{G: N_G \left({H}\right)}\right] \le \left[{G: H}\right]$

Each of the conjugates of $H$ has $\left\vert{H}\right\vert$ elements.

By Number of Distinct Conjugate Subsets is Index of Normalizer, there are $\left[{G: N_G \left({H}\right)}\right]$ conjugates.

Therefore:


 * $\left\vert{S}\right\vert \le \left\vert{H}\right\vert \left[{G: N_G \left({H}\right)}\right]$

with equality iff all conjugates of $H$ are disjoint.

Combining this with the earlier inequality, we find that:


 * $\left\vert{S}\right\vert \le \left\vert{H}\right\vert \left[{G: H}\right] = \left\vert{G}\right\vert$

Now if $H$ is normal in $G$, then by Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:


 * $\left[{G: N_G \left({H}\right)}\right] < \left[{G: H}\right]$

for $N_G \left({H}\right) = G$ in that case, and $H$ is a proper subgroup.

If $H$ is not normal in $G$, then it has multiple conjugates.

By Conjugate of Subgroup is Subgroup, they all are subgroups of $G$.

In particular, then, the identity of $G$ is common to all of them.

Hence they are not disjoint, and:


 * $\left\vert{S}\right\vert < \left\vert{H}\right\vert \left[{G: N_G \left({H}\right)}\right]$

In either case, it follows that:


 * $\left\vert{S}\right\vert < \left\vert{G}\right\vert$

as desired.