# Group, Cosets, Normal Group help

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• Sep 10th 2008, 08:27 PM
roporte
Group, Cosets, Normal Group help
Hello,

Could you help me in solving this problem?

---- Let G group, and H, K subgroups of G. Prove that if [G:H]=h and
[G:K]=k then lcm(h,k)|[G:H $\cap$K] <= h.k and if H or K are
normal subgroups of G then
[G:H intersection K] | h.k ----

I could prove that [G:H $\cap$K] <= h.k but i cannot prove
lcm(h,k)|[G:H intersection K] and if H or K are normal subgroups of G
then
[G:H $\cap$K] | h.k

Pleas i really appreciate very much your help and i'm sorry for my bad english

Cheers,
RP
• Sep 10th 2008, 08:52 PM
ThePerfectHacker
Quote:

Originally Posted by roporte
Hello,

Could you help me in solving this problem?

---- Let G group, and H, K subgroups of G. Prove that if [G:H]=h and
[G:K]=k then lcm(h,k)|[G:H $\cap$K] <= h.k and if H or K are
normal subgroups of G then
[G:H intersection K] | h.k ----

You prove that if $H,K$ have finite index then $H\cap K$ have finite index.
It follows that $H\cap K$ is finite index in $H$.
Now note $H\cap K \subseteq H \subseteq G$.
This means $(G:H\cap K) = (G:H)(H:H\cap K)$.
Therefore, $(G:H)|(G: H \cap K)$.
Now note $H\cap K \subseteq K\subseteq G$.
This means $(G:H\cap K) = (G:K)(K:H\cap K)$.
Therefore, $(G:K) | (G:H\cap K)$.
Thus, $\text{lcm}(h,k)$ must divide $(G: H\cap K)$.
• Sep 10th 2008, 08:59 PM
roporte
Thanks a lot!
• Sep 10th 2008, 09:12 PM
ThePerfectHacker
If $H,K$ are normal subgroups then $H\cap K$ is a normal subgroup. This means we can form the factor group $G/(H\cap K)$.

For the second part we wil use Lagrange's theorem. We will embed* $G/(H\cap K)$ in $G/H\times G/K$. Then it will follow that the index of $H\cap K$ in $G$ must divide $|G/H\times G/K | = hk$ and that would complete the proof. Note that $a(H\cap K) \mapsto (aH,aK)$ is an embedding (you need to show it is well-defined an a homomorphism).

*)Given a group $G_1$ and $G_2$ to embed $G_1$ in $G_2$ means to identity $G_1$ as an isomorphic subgroup of $G_2$ i.e. find a one-to-one homorphism $\theta: G_1\to G_2$. Then $G_1\simeq \theta (G_1)$ and $\theta(G_1)$ is a subgroup of $G_2$ because a homomorphic image of a subgroup remains a subgroup.