# Generic elementary group theory problems.

This question is about generic group theory problems. here are examples for what I’m referring to:

Prove that any group of order $p^2$, where $p$ is a prime, is abelian.

Let $G$ be a group of order $2n$. Suppose that half of the elements of $G$ are order $2$, and the other half form a subgroup $H$ of order $n$. Prove that $H$ is of odd order ($n$ is odd) and that it's abelian.

"Let $G$ be a group with $|G|=pq$ for some $p$, $q$ primes such that $p>q,q∤p−1$. Prove $G$ is cyclic."

Most of the time I find myself attacking these kind of problems with no coherent strategy, just throwing all my ammunition (Cauchy theorem, Lagrange theorem, index 2 theorem, intersection unions and multiplication of (normal-) subgroups etc.).

Best case scenario is I manage to prove the statement yet I don't quite get why the statement is true because the proof is so long and involves a lot of cases and assumptions by contradiction that i can't see the forest for the trees.

A lot of the times there are more than one way to prove the statement which are not so similar.

For an excellent example to what I’m referring to, look here.

Since I feel that most of the time I’m just juggling variables I'd like to know what should i do to understand what's really going on?

Another different thing that might ease my mind is an exhaustive list of the theorems that can be used to solve these kind of problems. That way i will at least know what are all the possible techniques that might work for these problems.

ADDED: Although neat proofs for the specific problems i posted are welcome they are not the reason i asked this question. I solved these problems and others too yet my proofs we’re long and lacked an identifiable idea. What I'm looking for is a general principal that could guide me in constructing proofs for these problems.

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Generic is way too much to ask for. For your first example knowng that the center of a finite $\;p$- group is non trivial and that the quotient $\;G/Z(G)\;$ cannot be cyclic non-trivial is enough. –  DonAntonio Dec 17 '13 at 17:13

I find that, for elementary group theory problems, one of the best places to start is by asking yourself if (how) groups actions could be used to express this. Then you have some very powerful (while still elementary) theorems that you can use.

For another example which was found in a homework problem in the abstract algebra class that I just finished:

Show that any group (including infinite) that contains a proper subgroup of finite index also contains a proper normal subgroup of finite index.

One way to prove it is to go through all of the motions of proving that the intersection of finite index subsets is, again, finite index, extending this by induction, looking at $N=\bigcap_{g\in G}gHg^{-1}$, proving that this normal and is actually a finite intersection by the Orbit-Stabilizer Theorem when $G$ acts on the cosets of $H$ by conjugation, and finally using the result on finite intersections of finite index sets that you would have proved.

The easier, and more powerful, way to show this result (and even more, as we'll see) is the following which comes from considering the action of $G$ on the cosets of $H$ by conjugation as the important feature, rather than just a means to show that the previous set $N$ is a finite intersection.

Let $H\leq G$ be of finite index $n$. Then, $G$ acts on the cosets of $H$ by conjugation, and this induces a map $\varphi:G\to\text{Sym}\left(G/H\right)\cong S_{n}$. The kernel of this map is normal, and $G/\ker\varphi$ isomorphic to a subgroup of $S_{n}$, so $|G:\ker\varphi|\leq n!$, as required.

This proves that, not only do we have a proper normal subgroup of finite index, but we have one of index $\leq n!$. So, group actions are very powerful and allow you to go straight to the core of many elementary group theory problems, rather than having you flounder around for a bunch of other results just to scratch the surface.

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Thanks! that was really helpful. I’ll try to work more with actions and see where that takes me. –  user116395 Dec 17 '13 at 21:36
Take note that these two proofs construct the same subgroup. I strongly agree that the latter one is a better way to think about it. Group actions are the foundation of almost everything of significance in group theory, including representation theory. –  User-33433 Dec 17 '13 at 23:28

I find both of these facts simplest to understand as one group acting on another (by conjugation).

In the first case: Once you know that a $p$-group cannot have trivial center, then the question is: in how many ways can a group of order $p$ act on another group of order $p$? Not many.

In the second, we have a group $K$ of order $2$ (pick one, any one) acting on a group $H$ of order $n$ (index $2$ subgroups must be normal), but in a very special way. The condition that everything outside of $H$ have order $2$ is equivalent to the statement that $K$ acts by inversion on $H$: $x\mapsto x^{-1}$. And when is this a homomorphism? Not often.

(The condition that $H$ has odd order is just a consequence of the fact that it cannot contain elements of order $2$.)

There are some simple statements—say, the Feit-Thompson theorem—that require very deep theory, so it is almost certainly impossible to give general rules for how to look at such things.

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Many times when one is stuck against a problem a good technique is to stop and try to think about the object you're trying to study, find out as much possible about that object. That's basically throwing every ammunition you have hoping to find the solution.

About the problem of understanding the reason why a fact is true: the point is that many time to really understand the deep reason for a fact you have to use concept that aren't the basic ones.

For instance for proving that every group of order $p^2$ is abelian one can use the fact that there's an element of order $p$ in the center. Of course you have to know this fact first and to prove it you have to use the class formula, which isn't one of the first fact you learn about groups. What I mean is that usually to understand deep reason for facts you have to known lot's of theory.

Hope this helps.

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First can I offer some encouragement: you said "I manage to prove the statement yet I don't quite get why the statement is true", well don't worry - we all feel like that a lot of the time!!! If you keep going in the subject and then come back to the more elementary results later you will often find you say "OHHH I get it now". This is much the same as what @Giorgio said.

One more practical suggestion I would like to make: when you have finished a proof, in some sense it's a huge psychological relief. You've been working at the problem for hours (days... months... years... a whole lifetime maybe) and you are so glad to have finished it that you just want to go on to the next thing. BUT if you stop, and read through the proof really critically - imagine you are trying to understand someone else's proof - you will often find that you are able to edit it. Maybe just make the explanations a bit clearer, maybe find a short cut, maybe realise that two cases are actually the same; perhaps this will start you thinking of a whole different approach. Slowly and bit by bit you can gradually get the proof into the best possible form. AND you will probably find your intuitive grasp of the problem increasing.

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Here's a proof of the first problem, which is certainly overkill. We use representation theory. We know that the size of G is the sum of the squares of the degrees of its irreducible representations. Moreover, if $\rho$ is an irreducible representation of $G$, then $deg(\rho)$ divides the order of the group, which is $p^2$. First off, we know that the group has the trivial representation, which is of degree one. If it had one irreducible representation of degree $p$ or larger, than automaticaly, the sum of the squares of the degrees of the irreducible representations would exceed the size of the group, which is a contridiction. Hence, it then follows that all irreducible representations of $G$ are of dimension 1. Thus, G is an Abelian group

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Solution to $2$:
To prove that $n$ is odd note that if $n$ is even then there exists Sylow $2$ subgroup $S$ of G, easy to see that exist half of elements in $S$ of order $2$ and other half is $S\cap H$ is subgroup of S, but now $|S|=2^k, k\geq 2$, so use induction.
To prove that $G$ is Abelian use: $\forall x\in G\setminus H$, $\forall a_1, a_2\in H$, $xa_2a_1\in G\setminus H$, so $(xa_2a_1)^2=e$, so $xa_2a_1=a_1^{-1}xa_2$, $(a_1^{-1}x)^2=e$, so $a_1^{-1}x=xa_1$ and $xa_2a_1=xa_1a_2$. done