Every group is normal in itself

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

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 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.