Subgroup is Normal iff it contains Product of Inverses
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Theorem
A subgroup $H$ of a group $G$ is normal if and only if:
- $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$
Proof
Necessary Condition
Suppose $H$ is normal.
Let $a b \in H$.
\(\ds a b\) | \(\in\) | \(\ds H\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^{-1} \left({a b}\right) a\) | \(\in\) | \(\ds H\) | Subgroup is Normal iff Contains Conjugate Elements | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b a\) | \(\in\) | \(\ds H\) | Group properties |
As $H$ is a group, then $x \in H \iff x^{-1} \in H$.
So $b a \in H \iff \left({b a}\right)^{-1} \in H \iff a^{-1} b^{-1} \in H$.
Thus $\forall a, b \in G: a b \in H \implies a^{-1} b^{-1} \in H$.
Sufficient Condition
Suppose $a b \in H \implies a^{-1} b^{-1} \in H$.
Then (from the above) we have $\forall a, b \in H: a b \in H \implies b a \in H$.
Let $g \in G$, and let $h \in H$. Let $h = x g$ where $x \in G$.
Now $g h g^{-1} = g x g g^{-1} = g x$.
By hypothesis, $x g \in H \implies g x \in H$.
The result follows.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 46 \beta$