Trivial subgroup is normal

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This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., trivially true subgroup property)
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Statement

Let G be any group. The trivial subgroup of G, which is the one-element subgroup comprising the identity element, is a normal subgroup of G.

Related facts

Proof

Conjugation definition of normality

Given: A group G with identity element e. An element g \in G.

To prove: geg^{-1} = e.

Proof: We have geg^{-1} = gg^{-1} = e, completing the proof. Further information: manipulating equations in groups

Kernel of homomorphism definition of normality

Given: A group G with identity element e.

To prove: There exists a homomorphism \varphi:G \to K for some group K such that the kernel of \varphi is precisely the subgroup \{ e \}.

Proof: Take \varphi to be the identity map (i.e., \varphi(x) = x for all x \in G). This clearly satisfies the conditions for a homomorphism: \varphi(gh) = gh = \varphi(g)\varphi(h), \varphi(g^{-1}) = g^{-1} = \varphi(g)^{-1}, and \varphi(e) = e.

The kernel of \varphi is defined as the set of elements that map to e. But since this is the identity map, the set is precisely \{ e \}, completing the proof.

Coset definition of normality

Given: A group G with identity element e.

To prove: For any g \in G, g \{ e \} = \{ e \} g.

Proof: Both are clearly the same as the singleton set \{ g \}, because ge = eg = g.

Union of conjugacy classes definition of normality

Given: A group G with identity element e

To prove: \{ e \} is a union of conjugacy classes.

Proof: In fact, \{ e \} is a single conjugacy class, because for any g \in G, geg^{-1} = e.

Commutator definition of normality

Given: A group G, with identity element e.

To prove: If T = \{ e \}, then [G,T] is contained in T.

Proof: [G,T] is the subgroup generated by elements of the form [g,e], where g \in G. But [g,e] = geg^{-1}e^{-1} = gg^{-1} = e. Thus, all elements of the form [g,e] equal e, so the subgroup generated is equal to T.