# 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 $G$.

## 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 $T$.