# Normal Subgroups

Let $$G$$ be an abelian group, the composition in $$G$$ being denoted multiplicatively. Let $$H$$ be any subgroup of $$G$$. If $$x$$ is an element of $$G$$, then $$Hx$$ is a right coset of $$H$$ in $$G$$ and $$xH$$ is a left coset of $$H$$ in $$G$$. Also $$G$$ is abelian, therefore we must have $$Hx = xH\,\,\,\forall x \in G$$. However, it is possible that $$G$$ is not abelian, yet it is possesses a subgroup $$H$$ such that $$Hx = xH\,\,\,\forall x \in G$$. Such subgroups of $$G$$ fall under the category of normal subgroups, and they are very important.

__Definition__

A subgroup $$N$$ of a group $$G$$ is said to be a normal subgroup of $$G$$ if for every $$x \in G$$ and for every $$n \in N$$, $$xn{x^{ – 1}} \in N$$.

From this definition we can immediately conclude that $$N$$ is a normal subgroup of $$G$$ if and only if

\[xN{x^{ – 1}} \subset N\,\,\,\,\,\,\,\forall x \in G\]

Nimra Shakoor

December 20@ 10:35 pmIf o(G)=6 and H≠K are subgroup of G each of order 2 then show that HK cannot be a subgroup of G. Show also that G cannot have two subgroups of order 3.