Normality satisfies transfer condition

This article gives the statement, and possibly proof, of a subgroup property satisfying a subgroup metaproperty
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Statement

Verbal statement

If a subgroup is normal in the group, its intersection with any other subgroup is normal in that subgroup.

Symbolic statement

Let $H \le G$ be a normal subgroup and let $K$ be any subgroup of $G$. Then, $H \cap K \triangleleft K$.

Property-theoretic statement

The subgroup property of being normal satisfies the transfer condition.

Definitions used

Normal subgroup

A subgroup $H$ of a group $G$ is said to be normal if for any $g \in G$ and $h \in H$, $ghg^{-1} \in H$.

Transfer condition

A subgroup property $p$ is said to satisfy transfer condition if whenever $H, K$ are subgroups of $G$ and $H$ has property $p$ in $G$, $H \cap K$ has property $p$ in $K$.

Related facts

Further facts

• Second isomorphism theorem: This result equates the quotient of the non-normal subgroup, by the intersection, with the quotient of the product of subgroups, by the normal subgroup. Specifically, it states that if $H$ is normal in $G$ and $K$ is any subgroup of $G$, we have $H/(H \cap K) \cong HK/K$.

Related metaproperties satisfied by normality

• Normality satisfies intermediate subgroup condition: The intermediate subgroup condition is weaker. It says that if $H \le K \le G$ are subgroups are $H$ is normal in $G$, then $H$ is normal in $K$.
• Normality satisfies inverse image condition: The inverse image condition is stronger. It says that the inverse image of a normal subgroup under a homomorphism is normal.

Other metaproperties satisfied by normality, that are somewhat related:

Proof

Hands-on proof

Given: A group $G$, a normal subgroup $H \triangleleft G$ and a subgroup $K \le G$

To prove: $H \cap K \triangleleft K$. In other words, we need to prove that given any $g \in K$ and $h \in H \cap K$, $ghg^{-1} \in H \cap K$.

Proof: Since $h \in H \cap K$, we in particular have $h \in H$. Since $H \triangleleft G$ (viz $H$ is normal in $G$), $ghg^{-1} \in H$.

But we also have that $g \in K$ and $h \in K$. Since $K$ is a subgroup, $ghg^{-1} \in K$.

Combining these two facts, $ghg^{-1} \in H \cap K$.

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, More info, Page 88, Exercise 24
• Topics in Algebra by I. N. Herstein, More info, Page 53, Problem 5