Trivial subgroup is normal

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

• Every group is normal in itself
• Isomorph-free implies normal
• Endo-invariance implies trivially true

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 $T$.