Abstract Algebra - group

[nebbish]

Abstract Algebra -- group

Show that in a group G of order $$p^2$$ any normal subgroup of order p must lie in the center of G.

I am pretty sure here that p is supposed to be a prime number, as that is the stipulation in preceding and later problems. However, the problem statement does not explicitly state that p is prime here. I think that's just an oversight.

Attempted solution:

Let x be an element of N, a normal subgroup of G of order p. If x isn't e, then o(x)=p, as
o(x) must divide o(N). Then N = (x). Let $$x^r$$ be an arbitrary element of N. Let g be an arbitrary element of G.
Then by the normality of N, $$gx^rg^{-1}=x^s$$ for some s.

Then $$gx^r=x^sg$$. If I could show r = s then I would be done.

 

[Dick]

Look up conjugacy classes.

 

[nebbish]

I didn't end up looking anything up (beyond what I already knew when I came across the problem) but in any case I believe I have solved the problem.

 

[Dick]

Really? How? I was tearing my hair out before I remembered the conjugacy class thing.

 

[nebbish]

how i did it

Really? How? I was tearing my hair out before I remembered the conjugacy class thing.

If I've got this figured correctly, it turns out that either there is an element of order p^2, in which case immediately the group is cyclic and abelian (therefore any subgroup is in the center); or, each element x in G other than e is in a normal subgroup of order p, namely, (x).
This is partly based on the preceding problem (which I did not post, but perhaps should have) which shows that indeed the group must have a normal subgroup of order p -- and, further, I showed in that problem that the normal subgroup was (x). Anyway, it seems to turn out (if I've done the work correctly) that if there is no element of order p^2 then every element of G is in one of these normal cyclic subgroups of order p. Probably these are what you meant by the "conjugation classes." Then a little manipulation shows that elements from different subgroups commute with each other (since the subgroups are cyclic they automatically commute internally) -- i.e., xy =yx regardless of whether x and y belong to the same normal subgroup. In fact, this goes a long way toward showing that the entire group is abelian, which was the next problem in the series.

By the way, I appreciate the fact you were "tearing your hair out" for my benefit -- even if you had reasons beyond helping me. On the other hand, I'm sure I would have found your advice about conjugation classes too vague if I hadn't solved the problem myself. But then I suppose I could have come back to you for a little more of a hint.

Also, if any of the above seems wrong to you I would appreciate hearing about it. Of course, I didn't provide you the gritty details on how I achieved some of these results.

 

[d_leet]

If I've got this figured correctly, it turns out that either there is an element of order p^2, in which case immediately the group is cyclic and abelian

This is false even if p is prime. Consider p=2, and the Klein 4 group. It is true, however, that a group pf order p^2 where p is prime is always abelian though, but the proof of this involves the class equation and hence congugacy classes, unless there is a simpler proof of this of which I am unaware.

 

[nebbish]

nearly a heart attack!

This is false even if p is prime. Consider p=2, and the Klein 4 group. It is true, however, that a group pf order p^2 where p is prime is always abelian though, but the proof of this involves the class equation and hence congugacy classes, unless there is a simpler proof of this of which I am unaware.

I'm not sure what exactly you're saying is false. The Klein 4 group's elements are of order 2 (other than e). My comment referred to an element of order p^2, which is 4 in this case. The Klein 4 group has no such element.

To be clear, I said EITHER there is an element of order p^2, OR...

Just presenting a group without an element of order p^2 is no counterexample at all.

On the other hand, perhaps you are saying that not every element of the Klein 4 group belongs to a cyclic normal subgroup of order 2. You would be wrong on that score as well. Every element of the Klein 4 group does belong to a cyclic normal subgroup of order 2. Therefore my statement is certainly true as it applies to the Klein 4 group. I will concede that I may have made an error and there is some other counterexample, but you have not presented one.

By the way, I appreciate the attention you are paying to my homework. Perhaps you will find a mistake. I hope it doesn't give me a heart attack when you do.

 

[d_leet]

Maybe I should read thingss more carefully, I missed the either in your sentence and did not initially read the rest of your explanation, so you can disregard my "counterexample".

 

[morphism]

While you may have solved this problem correctly, I second Dick's advice on looking up conjugacy classes and the class equation. A little understanding of the concepts behind them will go a long way.

 

[Dick]

morphism is right. The conjugacy class equation says that the order of a group is the sum of the order of its center plus the sum of the orders of all of the other conjugacy classes and the orders of those conjugacy classes must by divisors of the order of the group and >1. Since one plus something divisible by p can't divide p, this tells you you can't have a nontrivial center. The rest is gravy. It can save you a lot of work. I "tear my hair out" over a lot of easy problems. And, sure, it was JUST FOR YOU. :)

 

[nebbish]

you're right

While you may have solved this problem correctly, I second Dick's advice on looking up conjugacy classes and the class equation. A little understanding of the concepts behind them will go a long way.

Undoubtedly you are all correct and I hope we cover that material in short order. If not, I should go after it myself.

To a large extent I blame myself for not providing you guys with more information about how my textbook author intended the problem to be solved. This would have been clear if I had posted the problem preceding this one in the assignment, and the lemma which was intended to help us solve that problem. I plead ignorance. Really. I was too ignorant of the material to realize the significance of these factors. However, I will try to provide more of this type of background material in future requests for help.

Anyway, you guys rock.

P.S. Looking a little ahead in the book I see that indeed the author uses the class equation to do the proof as you suggest. However, this convinces me he did not intend for us to use the class equation "prematurely."