# Thread: Group theory subnormality question

1. ## Group theory subnormality question

How does one go about proving the subnormality intersection property, that is to say, if two subgroups are subnormal in G thier intersection is also?

2. Originally Posted by prettyboy
How does one go about proving the subnormality intersection property, that is to say, if two subgroups are subnormal in G thier intersection is also?
let H and K be two subnormal subgroups of G. let $\displaystyle H=H_0 \lhd H_1 \lhd \cdots \lhd H_m=G$ and $\displaystyle K=K_0 \lhd K_1 \lhd \cdots \lhd K_n=G$ be some subnormal series for H and K.

then $\displaystyle H \cap K=H_0 \cap K \lhd H_1 \cap K \lhd \cdots \lhd H_m \cap K=G \cap K=K \lhd K_1 \lhd \cdots \lhd K_n=G$ is a subnormal series for $\displaystyle H \cap K.$

3. Originally Posted by prettyboy
How does one go about proving the subnormality intersection property, that is to say, if two subgroups are subnormal in G their intersection is also?
Let $\displaystyle H,K \unlhd G$. Then for all $\displaystyle h \in H \cap K$ and all $\displaystyle g \in G$ we have that $\displaystyle g^{-1}hg \in H$ and $\displaystyle g^{-1}hg \in K$, so clearly $\displaystyle g^{-1}hg \in H \cap K$, and so $\displaystyle H \cap K \unlhd G$ holds.

For more practice with the intersections of groups, can you prove that the intersection of finitely many subgroups of finite index is a subgroup of finite index?

4. Originally Posted by Swlabr
Let $\displaystyle H,K \unlhd G$. Then for all $\displaystyle h \in H \cap K$ and all $\displaystyle g \in G$ we have that $\displaystyle g^{-1}hg \in H$ and $\displaystyle g^{-1}hg \in K$, so clearly $\displaystyle g^{-1}hg \in H \cap K$, and so $\displaystyle H \cap K \unlhd G$ holds.
H and K are subnormal not normal. see the question again.

5. Originally Posted by NonCommAlg
H and K are subnormal not normal. see the question again.

Oooh-I did wonder what you were going on about in your post!

That said, it is just a logical extension of the fact that I proved and so it's not an entirely irrelevant proof...