Trivial subgroup is normal

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

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.

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 $$T$$.