Normality satisfies transfer condition

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

Other metaproperties satisfied by normality, that are somewhat related:

Transfer condition for other subgroup properties

Analogues in other algebraic structure


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.


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