Direct factor implies central factor

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., direct factor) must also satisfy the second subgroup property (i.e., central factor)
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Statement

Verbal statement

Any direct factor of a group is a central factor.

Statement with symbols

Suppose $H$ is a direct factor of a group $G$, i.e., $H$ is a normal subgroup of $G$ and there exists a normal subgroup $K$ of $G$ such that $HK=G$ and $H\cap K$ is trivial. Then, $H$ is a central factor of $G$, i.e., $H{C}_G(H)=G$.

Related facts

Converse

• Central factor not implies direct factor
• Central factor not implies complemented
• Complemented central factor not implies direct factor

Stronger facts

• Direct factor implies right-quotient-transitively central factor
• Direct factor implies join-transitively central factor
• Join of finitely many direct factors implies central factor

Other related facts

• Image-potentially direct factor equals central factor
• Internal direct product implies internal central product
• Direct factor implies transitively normal
• Central factor implies transitively normal

Facts used

1. Internal direct product implies internal central product

Proof

Given: A group $G$, normal subgroups $H,K$ of $G$ such that $HK=G$ and $H\cap K$ is trivial.

To prove: $H{C}_G(H)=G$.

Proof:

1. Every element of $H$ commutes with every element of $K$: For $h\in H$ and $k\in K$, the commutator $[h,k]=hk{h}^{-1}{k}^{-1}$ is in $H$ (because $H$ is normal) and is also in $K$ (because $K$ is normal). (This is based on one of the equivalent definitions of normal subgroup. It can also be seen by seeing that $[h,k]=h(k{h}^{-1}{k}^{-1})=(hk{h}^{-1})k^{-1}$). Since $H\cap K$ is trivial, we obtain that $[h,k]$ is the identity element, so $hk=kh$.
2. $K\le C_G(H)$: This is a reformulation of the previous step.
3. $H{C}_G(H)=G$: Since $K\le C_G(H)$, $G=HK\le H{C}_G(H)\le G$. Equality holds throughout, so $H{C}_G(H)=G$.