Could you check the problem statement? As it stands, the statement is false. For example, the Icosahedral group is a finite simple group.
Hey I have a question out of my algebra text (Hungerford, p.92 #13)
It asks to show that if G is a group containing a proper subgroup H of finite index, then it contains a proper normal subgroup of finite index.
It's easy to show the fact if G is finite or abelian, but I'm left with the last case if G is nonabelian and infinite. I'm trying to work with the center, which I know is normal, but I don't know how to manipulate it to create a resultant normal group of finite index, and obviously I need to incorporate the given subgroup H somehow.
Letting C be the center of G, I tried to consider the group HC = { hc | h in H, c in C} but I get no success trying to show this is either normal or has finite index. I am terrible with conjugation in groups in general, and considering this exercise involves a degree of counting cosets, I'm suspecting conjugation is going to play a part, which is even more intimidating.
Any assistance would be appreciated.
The proposition holds as stated, even for simple groups. The proper normal subgroup induced doesn't have to be nontrivial. The group generated by the identity is normal and has index equal to the order of the group, which is finite for finite groups.
In the meantime, I sought help through my professor, and we established a homomorphism between G and the symmetric group of H's left cosets. The kernel of this is the subgroup I've been looking for: it's obviously normal, and it has finite index because G/Ker can't have more than n! elements, where n is the index of H.
Thanks anyway. I'm sure I'll have more.
Right, the proposition doesn't guarantee the induced normal subgroup is necessarily nontrivial. In , the subgroup generated by the identity permutation is normal and isn't equal to , so it satisfies the conclusion. The fact that it's trivial doesn't contradict the proposition.
For finite groups (with at least two elements), the proposition itself is easy because the trivial subgroup is always normal and has finite index. The challenge I had was showing the statement for nonabelian infinite groups, in which the trivial subgroup has infinite index so I had to find another one.
Here is what your problem comes down to:
"If G is infinite and has a proper group of finite index it cannot be simple*."
Look at the contrapositive,
"If G is infinite and has a proper subgroup of finite index it cannot have be simple".
Let us prove the contrapositive by contradiction. Let G be and infinite group and H < G where (G:H) is finite. We can view G as an G-set by defining a group action on H as "left-translation on the cosets of H"**. Let X be the left cosets of H then this creates a homomorphism f between G and S_X (symettric group). Thus, there is a homomorphic map f: G --> S_X.*** This mapping cannot possibly be one-to-one because G is an infinite group and |S_n|=n!. So kernel(f) cannot be trivial. Since G is simple it means kernel(f) = G. But H contains kernerl(f), thus H=G. We have shown that if (G:H) is finite it means H=G, i.e. G has no proper subgroups of finite index. Q.E.D.
*)It has not proper non-trivial normal subgroups.
**)Define the operation g(aH) = (ga)H, this maps left cosets of H into left cosets of H and satisfies the conditions for being a G-set.
***)This should be known to you. If not define f(g) = i_g(x) where i_g(x) is the map i_g(x) = gx. Now show it is a homomorphism.
I can retype this in LaTeX if you want, but group theory is often were simple on notation.