Direct factor implies central factor

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., $$HC_G(H) = G$$.

Converse

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

Stronger facts

 * Weaker than::Direct factor implies right-quotient-transitively central factor
 * Weaker than::Direct factor implies join-transitively central factor
 * Weaker than::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) uses::Internal direct product implies internal central product

Proof using the product with centralizer definition
Given: A group $$G$$, normal subgroups $$H,K$$ of $$G$$ such that $$HK = G$$ and $$H \cap K$$ is trivial.

To prove: $$HC_G(H) = G$$.

Proof:


 * 1) Every element of $$H$$ commutes with every element of $$K$$: For $$h \in H$$ and $$k \in K$$, the commutator $$[h,k] = hkh^{-1}k^{-1}$$ 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 $$[h,k] = h(kh^{-1}k^{-1}) = (hkh^{-1})k^{-1}$$). 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) $$HC_G(H) = G$$: Since $$K \le C_G(H)$$, $$G = HK \le HC_G(H) \le G$$. Equality holds throughout, so $$HC_G(H) = G$$.

Note that step (1) above is sometimes taken as part of the definition of internal direct product, in which case it does not need to be proved.

Hands-off proof using direct product and central product
By fact (1), an internal direct product is an internal central product. A direct factor is a factor in an internal direct product, and a central factor is a factor in an internal central product. Thus, a direct factor must be a central factor.

Proof using the inner automorphism definition of central factor
Given:A group $$G$$ that is an internal direct product of subgroups $$H$$ and $$K$$. In other words, $$HK = G$$ and $$H \cap K$$ is trivial.

To prove: For any $$g \in G$$, there exists $$h \in H$$ such that, restricted to $$H$$, conjugation by $$g$$ equals conjugation by $$h$$.

Proof: We can write $$g = ab$$ with $$a \in H$$, $$b \in K$$ since $$G = HK$$.


 * 1) Every element of $$H$$ commutes with every element of $$K$$: For $$h \in H$$ and $$k \in K$$, the commutator $$[h,k] = hkh^{-1}k^{-1}$$ 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 $$[h,k] = h(kh^{-1}k^{-1}) = (hkh^{-1})k^{-1}$$). Since $$H \cap K$$ is trivial, we obtain that $$[h,k]$$ is the identity element, so $$hk = kh$$.
 * 2) Restricted to $$H$$, conjugation by $$a$$ equals conjugation by $$g$$: If $$c_g, c_a, c_b$$ denote conjugation by $$g,a,b$$ respectively, then ,math>c_g = c_a \circ c_b . Restricting to $$H$$, $$c_b$$ is trivial by the previous step. Thus, $$c_g = c_a$$.