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Every way to confirm a subgroup is normal

Four shortcuts, cheapest first -- run the full conjugation test only when none of them apply.

Check the free cases first: is G abelian, is H a kernel, or is H the center? If none apply, check whether [G:H]=2 -- that alone settles it. Only when every shortcut misses do you need to test conjugation by a generating set, and only when even that isn't available do you fall back to checking ghg^{-1} \in H by hand.

Automatic, with zero conjugation checked

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A group in which the operation commutes: a*b = b*a for every pair of elements.

Example

\mathbb{Z}_6 is abelian, so its subgroup \{0,3\} is normal with no check at all.

In an abelian group every element commutes with every other, so gHg^{-1}=H holds for every subgroup automatically.

Term

kernel

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For a homomorphism f: G \to H, the set \ker(f) = \{g \in G : f(g) = e_H\} -- always a normal subgroup of G.

The kernel of any homomorphism is a normal subgroup of the domain, built in for free by the definition of a homomorphism.

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The set Z(G) of elements that commute with every element of G -- always a normal subgroup.

Example

Z(D_4)=\{e,r^2\} -- confirmed by testing every candidate against every g\in D_4.

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Z(G) consists exactly of the elements that commute with everything, so conjugating it by any g changes nothing.

Cheap shortcuts before running the full test

Normality tests →

Let H \le G with [G:H] = 2. Then H \trianglelefteq G.

Example

A_3 has index 2 in S_3 (|S_3|/|A_3| = 6/3 = 2), so A_3 \trianglelefteq S_3 with a single division, no conjugation needed.

Whenever H accounts for exactly half of G, there's only one leftover clump outside it, so gH and Hg always land on the same clump.

Show proof

Direct, using left and right cosets. Since [G:H]=2, G splits into exactly two left cosets. One of them is H itself (eH=H); since the two left cosets partition G and are disjoint, the other one must be exactly the complement G \setminus H. The identical argument for right cosets shows G also splits into exactly the two right cosets H and G \setminus H.

Now check gH=Hg for every g \in G. If g \in H, then gH = H = Hg (closure). If g \notin H, then gH is the OTHER left coset (it cannot equal H, since g \in gH but g \notin H), so gH = G \setminus H -- and by the identical reasoning, Hg = G \setminus H too. Either way gH = Hg, so H \trianglelefteq G.

The general test, when nothing above applies

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A subgroup H \le G with gHg^{-1}=H for every g \in G -- equivalently, gH=Hg for every g.

Example

D_4's rotation subgroup R=\{e,r,r^2,r^3\}: conjugating by r is automatic (r\in R), and srs^{-1}=r^{-1}=r^3\in R handles the one remaining generator.

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Check ghg^{-1}\in H for every g\in G and h\in H -- checking this one-sided inclusion for every g is already enough; the reverse inclusion comes free.

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