Permutation of Cosets/Corollary 2

Corollary to Permutation of Cosets

Let $G$ be a group.

Let $p$ be the smallest prime such that:

$p \divides \order G$

where $\divides$ denotes divisibility.

Let $\exists H: H \le G$ such that $\order H = p$.

Then $H$ is a normal subgroup of $G$.

Proof

Apply Permutation of Cosets: Corollary 1 to $H$ to find some $N \lhd G$ such that:

$\index G N \divides p!$

Because $\index G N \divides \order G$, it divides $\gcd \set {\order G, p!}$.

Because $p$ is the smallest prime dividing $\order G$, it follows that:

$\gcd \set {\order G, p!} = p$

Thus:

$\index G N = p = \index G H$

Because $N \subseteq H$, it must follow that $N = H$.

$\blacksquare$

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $9$: Permutations: Corollary $9.25$