Every group is normal in itself

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

Statement with symbols

Suppose $G$ is a group. Consider $G$ as a subgroup of itself -- $G$ is a normal subgroup of itself.

Related facts

Opposite facts

• Subgroup isomorphic to whole group need not be normal

Similar facts

• Index two implies normal
• Subgroup of least prime index is normal

Other related facts

• Invariance implies identity-true
• Balanced implies identity-true
• Every group is characteristic in itself
• Trivial subgroup is normal

Proof

Proof using the conjugation definition of normality

Given: A group $G$.

To prove: For any $g\in G$ and $h\in G$, $gh{g}^{-1}\in G$.

Proof: This is obvious from the fact that $G$ is closed under multiplication.

Proof using the kernel of homomorphism definition of normality

Given: A group $G$.

To prove: There exists a homomorphism $\phi :G\to K$ for some group $K$ such that every element of $G$ maps to the identity element.

Proof: Let $K$ be the trivial group and $\phi$ be the map sending every element of $G$ to the identity element of $K$. Clearly, $\phi$ satisfies the conditions for being a homomorphism: for any $g,h\in G$, both $\phi (gh)$ and $\phi (g)\phi (h)$ equal the identity element of $K$. Moreover, every element of $G$ is sent to the identity element of $K$ under $\phi$.

Proof using the cosets definition of normality

Given: A group $G$.

To prove: For every element $g\in G$, $gG=Gg$.

Proof: Note that

1. $gG=G$: Clearly, $gG\subseteq G$. Also, for any $h\in G$, $h=g(g^{-1}h)\in gG$, so $G\subseteq gG$. Thus, $gG=G$.
2. $Gg=G$: Clearly, $Gg\subseteq G$. Also, for any $h\in G$, $h=(h{g}^{-1})g\in Gg$, so $G\subseteq Gg$. Thus, $Gg=G$.

Combining the two steps, we obtain that $Gg=gG$.

Proof using the union of conjugacy classes definition of normality

Given: A group $G$.

To prove: $G$ is a union of conjugacy classes in $G$.

Proof: The conjugacy classes form a partition of $G$ (arising from the equivalence relation of being conjugate), so $G$ is their union.

Proof using the commutator definition of normality

Given: A group $G$.

To prove: For every $g,h\in G$, $[g,h]\in G$.

Proof: This is direct from the fact that $G$ is closed under multiplication and inverses, and the commutator is defined in terms of these operations.