# Subgroup is Normal iff Left Cosets are Right Cosets

## Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $N$ be a subgroup of $G$.

Then $N$ is normal in $G$ (by definition 1) if and only if:

- Every right coset of $N$ in $G$ is a left coset

or equivalently:

- The right coset space of $N$ in $G$ equals its left coset space.

## Proof

### Necessary Condition

Let $N$ be a normal subgroup of $G$ by Definition 1.

Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.

$\Box$

### Sufficient Condition

Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.

Let $g \in G$.

Since every right coset of $N$ in $G$ is a left coset, there exists an $h \in G$ such that $N \circ g = h \circ N$.

By Element of Group is in its own Coset:

- $g \in N \circ g = h \circ N$

From Element in Left Coset iff Product with Inverse in Subgroup:

- $g^{-1} \circ h \in N$

Then:

\(\displaystyle N \circ g\) | \(=\) | \(\displaystyle \paren {g \circ g^{-1} } \circ \paren {h \circ N}\) | $\quad$ Definition of Inverse Element and Coset by Identity | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g \circ \paren {\paren { g^{-1} \circ h } \circ N}\) | $\quad$ Subset Product within Semigroup is Associative: Corollary | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle g \circ N\) | $\quad$ $g^{-1} \circ h \in N$ and Left Coset Equals Subgroup iff Element in Subgroup | $\quad$ |

Since this holds for all $g \in G$, $N$ is normal in $G$ (by definition 1).

$\blacksquare$

## Also see

## Sources

- 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \alpha$ - 1996: John F. Humphreys:
*A Course in Group Theory*... (previous) ... (next): Chapter $7$: Normal subgroups and quotient groups: Proposition $7.4 \ \text{(e)}$