Central implies normal

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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., central subgroup) must also satisfy the second subgroup property (i.e., normal subgroup)
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Suppose H is a subgroup of G that is a central subgroup of G, i.e., H is contained in the center Z(G) of G. Then, H is a normal subgroup of G.

Related properties

Intermediate properties

Property Proof that central subgroup implies it Proof that it does not imply central Proof that it implies normal subgroup Proof that normal subgroup does not imply it
central factor central implies central factor central factor not implies central central factor implies normal normal not implies central factor
hereditarily normal subgroup central implies hereditarily normal hereditarily normal not implies central (immediate) (via transitively normal)
transitively normal subgroup (via hereditarily normal) (via hereditarily normal) (immediate) normality is not transitive
abelian normal subgroup central implies abelian and central implies normal abelian normal not implies central (immediate) every group is normal in itself, combined with existence of non-abelian groups.

Here's a more complete list: Abelian normal subgroup, Amalgam-characteristic subgroup, Amalgam-strictly characteristic subgroup, Center-fixing automorphism-invariant subgroup, Class two normal subgroup, Commutator-in-center subgroup, Conjugacy-closed normal subgroup, Dedekind normal subgroup, Direct factor over central subgroup, Hereditarily normal subgroup, Join-transitively central factor, Nilpotent normal subgroup, SCAB-subgroup, Transitively normal subgroup|FULL LIST, MORE INFO

Related facts


The converse is not true in general: normal not implies central.

However, some versions are true:


Proof using coset definition of normality

Given: A group G, a central subgroup H of G.

To prove: gH = Hg for all g \in G (the cosets definition of normality).

Proof: Since H is central, this means that gh = hg for all g \in G, h \in H. Thus, for a fixed g, the sets gH = \{ gh: h \in H \} and Hg = \{ hg : h \in H \} are equal, because each gh equals the corresponding hg.

Proof using conjugation definition of normality

Given: A group G, a central subgroup H of G.

To prove: For all g \in G and h \in H, we have ghg^{-1} \in H.

Proof: Since h is central, we have, by definition, that gh = hg for all g \in G. Multiplying both sides on the right by g^{-1}, we obtain that ghg^{-1} = h for all g \in G. Since h \in H by assumption, and ghg^{-1} = h, we obtain that ghg^{-1} = h.

Proof using commutator definition of normality

Given: A group G, a central subgroup H of G.

To prove: For all g \in G, h \in H, we have ghg^{-1}h^{-1} \in H.

Proof: Since h \in H we have gh = hg by definition. Multiplying both sides by g^{-1}h^{-1} on the right, we get ghg^{-1}h^{-1} = e (i.e., it is the identity element). Since any subgroup contains the identity element, e \in H, so Failed to parse (syntax error): ghg^{-1}{h^{-1} \in H .

Proof using union of conjugacy classes definition of normality

Given: A group G, a central subgroup H of G.

To prove: H is a union of conjugacy classes in G.

Proof: Every element of the center of G forms a conjugacy class of size 1. Since H comprises only central elements, it is the union of these singleton conjugacy classes.