Groups/normal subgroups

  • Thread starter calvino
  • Start date
  • #1
i have the following problems that i can't work out. any help would be appreciated.

Problem 1: Let G be a group generated by elements a, b with a^4 = e, b^2 = e, baba = e. Show that G has at most 8 elements. Is there such a group with precisely 8 elements?

Problem 2: Let H, K be normal subgroups of a group G whose intersection contains only e_G, and whose union generates the whole group G. Prove that G is isomorphic to H x K. (Hint: Prove that hk = kh for all h in H and k in K, and consider the map H x K --> G with (h,k) |--> hk.)

with 1, it's easy to show that G has at most 8 elements. I'd simply write them all out, and come across some congruent elements. So we'd have the following group

bb=e -----> shows b cannot generate anymore distinct elements

However, is there a group with exactly 8 elements? Thinking about it...we'd have to have exactly one representation of each value. If each is to have it's own inverse, then thered have to be an even amount of elements, excluding the neutral. This means that either an odd amount of elements are their own inverses, or that theyre all all theyre own inverses. I believe that it doesn't work out , but how can i prove it? Any help as for where to go?

As for not sure where to begin either, even though the hint is there. I will keep you posted with my progress however. thank you in advance
  • #2
For the first, how do you know ab=ba? If that was true, then you could always take an arbitrary element (like aababbabbab) and rearrange it into the form anbm, and then reduce n mod 4 and m mod 2. But I don't think it's true that ab=ba.

For the second, do you understand the clue? It maps out how to do the problem, so which part are you getting stuck at?
  • #3
you're completely right with 1)... for some reason i skipped entirely over the fact that baba=e. it's still easy that way. thanks for the observation. any clue as to how i can show there is no group of order 8, strictly with 8 elements?

as for 2), i guess i have trouble with proofs involving normal subgroups. N and H are normal in G, so by definition gng^-1 is in N, and ghg^-1 is in H for all n in N, h in H, and g in G. So i look at the entire subgroups NH, which is simply the product of nh's. I examine the possibilities gNH=NHg=NgH, etc. and i get stuck. I know I'm suppose to utilize e_G, and the fact that their (N and H) unions generate G, but I am not sure how. I guess that by using those facts we can somehow get commtativity? I am currently still analyzing it. so ill keep you updated.
  • #4
actually i think i get a little more of it...since H is normal in G, gH=H let g be an element not in H. So it in N or generated by n in N, which would mean it is N anwyay (right?).

Because of normality,
gH=Hg, and using this, we can somehow manipulate to get nh=hn for all n in N and h in H

***SORRY...when i speak of N and n, i mean K, and k respectively. ***
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  • #5
For 1, there is such a group, and you still haven't shown it can have at most 8 elements.

For 2, I don't understand what you've done. gH is not the same as H unless g is in H. Start as they've suggested by showing hk=kh. The trick is to try to form an equality between an element in H and one in K, and then use the fact that their intersection is {e} to show this element must be e.
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  • #6
for 1)
considering, baba=e. we can take inverses of a and b to get

babaa^-1= e a^-1
which implies bab= a^-1.

Similarly aba= b^-1.

Furthermore we get ab=b^-1 a^-1
and ba= a^-1 b^-1.

By continuing on like so, which I plan to do, I'm pretty sure that I can show it has at most 8 elements.

Let me start typing what I think of 2), in another post.
  • #7
so let g be an element of G not in H.

gH is a subset of K.

so gh=k for some k in K, h in H

Edit: i originally started talking about another element g' in G and K such that g'k'=k also (fomr some k' in K). I didnt know, nor do I now, how to continue.
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  • #8
gH is not a subset of K. The union of H and K generates G, it isn't G by itself, and you can have elements that are in neither H nor K.

You're trying to show h commutes with k, so you need to show the commutator is e.
  • #9
oh sorry...i never used a theorem involving commutator. looked it up now. i'll work on it and reply in a few hours...thanks for your help.

edit: i don't think we can use theorems nor definitions that haven't been taught in class. is there another way to prove the commumativity?
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  • #10
It's not a theorem really, all you need is that if hkh-1k-1=e, then hk=kh, but this just comes from multiplying both sides by kh on the right. Can you show that hkh-1k-1=e? (remember what the definition of a normal group means)
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  • #11
thanks..i think i have it. looking over 1 though, today, I'm confused with the "at most" elements, and finding the group with exactly 8 elements. firstly, i thought that I would be able to find all 8 elements needed for part a, and that there would be 8. second of all, does part b mean that there exists a group with 8 elements and no congruencies? it seems as if I can find a congruency in any group.
  • #12
just to show my progress in 2)..

what i did was show that the commutator was an element in K, and that it is in H, using the definition of Normal in G. this made it e. thanks for the hints, though I am sure you might think you've overhinted. i appreciate it though. ill keep you all updated with how i finish off.
  • #13
ok so i showed that the function given in the hint was a homomorphism (straight from definition), and now I'm just going to show that it is 1-1, by proving something about the kernel. thanks for the help again.

now, as for 1) unsure of how or which group has EXACTLY 8 elements. like i said, i keep believing each value can have a congruency somehow.
  • #14
I'm guessing you've shown there are 8 expressions (things like e,a,b,ab, etc), such that any product of a's and b's can be reduced to one of these 8 expressions, and so there are at most 8 elements in the group. So just assume that all 8 expressions are distinct elements and show this gives you a group.
  • #15
Here's a hint:
If, the group is commutative (as you originally assumed) you have:
and from the problem you have:
Then, assuming [itex]a \neq e[/itex], the group generated by [itex]a[/itex] and [itex]b[/itex] has the elements:
  • #16
The group of part 1 D_4 (some authors call it D_8). The symmetries of the square. i.e. 1-1 and onto mappings from the square to itself. It is a subgroup of S_4 generated by the cycle (1234) and the transposition (13)
  • #17
That is only true if you assume that the equations given are "relations", i.e. a=/=b and no smaller powers of a or b are the identity (or the order of a=4 and the order of b=2, if I recall the previous page correctly).
  • #18
yes, i figured out that the symmetry of the square did work. however, for the first part, i messed up and somehow assumed ba=ab. the way i attempted to prove it was by claiming certain elements were the generators of the set, and by trying to show just that. obviously getting ab=ba (somehow!) gave me the wrong generators. was that the right technique though? because it seems that would prove that the group had exaxctly 8, and not less or equal to 8.

i tried looking at inclusion/exclusion afterwards, since |<a>|=4, |<b>|=2, |<ab>|=2, |<a^-1>|=4 and every element had to be a product of a,b, or a^-1. I had thought this would get me 8 - |some intersection|, but I got stuck. So I still think that the suggestion or idea i have in the first paragraph is the way to go.
  • #19
You know there are at most 8 elements of the form aibj, since you can reduce i (mod 4) and j (mod 2). Can you show all elements in the group can be written in this form?

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