Internal direct product

A way of recognising that a group $G$ is internally built up from two of its subgroups.

Definition

Let $G$ be a group with Normal subgroup $M,N⊴G$ such that

$$M\cap N=\{e\}\text{and}MN=G.$$

Then $G$ is the internal direct product of $M$ and $N$.

Equivalent: $G\cong M\times N$ Externally

Theorem

If $M,N⊴G$ with $M\cap N=\{e\}$, then $MN\le G$ and

$$MN\cong M\times N$$

(the external direct product).

So internal and external direct products are isomorphic; the distinction is whether you build $M\times N$ from outside (taking pairs) or detect it inside an existing group $G$ (finding two normal subgroups with trivial intersection).

Detection Recipe

To show $G\cong M\times N$:

1. Find $M,N⊴G$.
2. Verify $M\cap N=\{e\}$ (often via Lagrange when $|M|$ and $|N|$ are coprime).
3. Verify $MN=G$ (often via $|MN|=|M||N|/|M\cap N|=|G|$).

Then $G\cong M\times N$ follows automatically.

Example: Order $15$ Group is Cyclic

$|G|=15=3\cdot 5$. By Sylow’s theorems, $G$ has unique (hence normal) Sylow 3- and Sylow 5-subgroups $J,K$. $J\cap K=\{e\}$ by Lagrange ($\gcd (3,5)=1$). $|JK|=3\cdot 5=15=|G|$, so $JK=G$. Hence

$$G\cong J\times K\cong C_3\times C_5\cong C_{15}.$$

Generalisation to More Subgroups

If $M_1,\dots ,M_k⊴G$ have pairwise trivial intersections (and the right product condition), then $G\cong M_1\times \cdots \times M_k$. In particular, every finite abelian group decomposes as a direct product of cyclic prime-power groups (the Fundamental Theorem of Finite Abelian Groups).

Key Lemma

The technical heart of the theorem: if $M,N⊴G$ with $M\cap N=\{e\}$, then $mn=nm$ for all $m\in M$, $n\in N$ - they commute. Proof: $n^{-1}mn{m}^{-1}\in M\cap N=\{e\}$ via the commutator argument.

This commuting is what makes the multiplication in $MN$ behave like componentwise multiplication in $M\times N$, giving the isomorphism $\theta :MN\to M\times N$, $\theta (mn)=(m,n)$.