Group theory Dummit and Foote, Sec 3.2 no.10

[buoyant forge]

Suppose H and K are subgroups of finite index in the group G with
|G : H| = m and IG : K| = n.

Prove that
lcm(m, n) <=
|G : intersection of H and K| <= mn.

I was able to prove the right inequality by the formula |HK|=|H||k|/| the intersection of H and K |.
Can anyone give me a hint about how to solve the left one? Thanks!
Ps. Sry for not using latex , I'm new here...

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[ember tendon]

Prove that $n$ divides $[G:H \cap K]$ and $m$ divides $[G:H \cap K]$

[frosty fiber]

Rotor 🙂

[ember tendon]

Generally you can prove that if $J$ is a subgroup of $D$ and $D$ a subgroup of $S$ then $[S:J]=[S:D][D:J]$ such that $J$ has finite index in S

(Try proving that)

[frosty fiber]

Rotor 🙂

[ember tendon]

There are easier proofs of this in the case that G is finite(which isn’t necessarily the case) , or if you assume H and K to be normal then you can create a surjective group homomorphism between G/H inter K and G/H, and G/H inter K and G/K

[buoyant forge]

Ohhhh , that's simpler than I thought!

[buoyant forge]

I think another way to prove it is by using |H inter K| = |HK|/|H||K| ?

[ember tendon]

Do you mean |H inter K|=|H||K|/|HK|? (you always have |H| |K|>= |HK| if both are finite because you have a natural surjection (h,k)—->hk) The problem is that we don’t know if H or K are of finite order we just know that they have finite index

Like det:On(R)—->{-1,1} is a surjective group homorphism with kernel SOn(R) so using the first isomorphism theorem On(R)/SOn(R) ~{-1,1} so [On(R):SOn(R)]=2 while SOn(R) is not finite

[buoyant forge]

Yes yes , I noticed that 😢. I'm just so dumb. Well thanks for helping me !! You're such an lifesaver!!

[ember tendon]

Np, buy No you aren’t dumb you Just made a small mistake it’s no biggie

[buoyant forge]

Thank you! 🙏

+close

[gleaming wolf]

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