# 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.

## Proof

### Proof using the conjugation definition of normality

Given: A group $G$.

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

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 $\varphi:G \to K$ for some group $K$ such that every element of $G$ maps to the identity element.

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

### 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 $h \in G$, $h = (hg^{-1})g \in Gg$, 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.