If G is a group, and if If N1, N2, N3..... etc are Normal Sub-Groups of G.
Q1. How do I prove that the largest of the above will contain all the rest?
Thanks,
Aman
You can't possibly prove this because it isn't true. I don't know an easy finite counter-example off-hand, though I guess any you could do something like pick the cyclic Abelian group of order (30) which has normal subgroups with 5 3 and 2, elements which all have trivial intersection.
A nice infinite counter example comes from the wallpaper groups. P4MM contains C2MM P2MM and P4 as subgroups of index 2. They are all obviously normal subgroups but distinct from one another.
Well in the second case there certainly cannot be any larger subgroups. All the subgroups I gave are subgroups of index 2, therefore they are normal, and therefore no larger proper subgroup contains any of them. And my C(30) example works as well if we pick different subgroups. It clearly contains subgroups of index 2 and 3 that are both normal, and the index 2 subgroup does not contain the index 3 subgroup.
You are definitely trying to prove something that is not true, unless you want to make what you are trying to prove trivially true by noting that every group contains itself as a normal subgroup of index 1 and this obviously contains all the other normal subgroups.
Maybe you should post whatever question you have been given in full.
alunw, on second thoughts, by definion of normal group, G itself is normal. so infact my question is redundant.
maybe this is a better way to put my question
G is a group. H is a sub-group. Let N1, N2, N3 .. etc be normal sub-groups of G and each is a subset/subgroup of H.
Then how do we prove that the largest of above (which is again is normal sub-group of G and subset of H) contains all the rest.
Thanks
In the first place you need to specify that N1,N2,etc is the collection of ALL normal subgroups of G that are subgroups of H because if not I can certainly pick specific collections that will disprove your claim. Also you need to decide whether H is a proper subgroup of G, and whether the word "proper" should be inserted before "subgroups of H" when limiting the collection. Presumably it is not given that H itself is a normal subgroup of G.
Then you might have a conjecture which it would take me some time to come up with a counter-example for, and if it took a really long time, I might start to believe it could be true, but, since my knowledge of group theory is more practical than theoretical I would probably have a lot of trouble proving it.
What grounds do you have for believing your claim? If it is a known theorem then please provide a reference.
Sorry for my vague definitions, really new to the subject. All the points you mentioned are valid.
However I found the reasoning which might prove the above conjecture. If N1 and N2 are normal subgroups of G (which are subsets of H as well) then N1xN2 will also be normal. Also N1xN2 is a subgroup of H (it might be H itself in which case H is a normal subgroup and it is this normal sub group which contains all).
Hope this argument makes sense.
It does seem convincing if I am following what I think you are saying. Perhaps we should limit ourselves to finite groups before we get ahead of ourselves.
So let $\displaystyle \{N_1,N_2,...,N_k\}$ be a collection of normal subgroups such that $\displaystyle N_i \subset H \leq G$ for all $\displaystyle i$.
Then your suggestion is to consider $\displaystyle N^*=\prod_iN_i=\{n_1n_2\cdots n_k | n_i\in N_i\}$. It is clear $\displaystyle N\subset H$ and furthermore for any $\displaystyle n_i\in N_i$, the element $\displaystyle 1\cdot1\cdots n_i \cdots 1 \in N$, so $\displaystyle \cup N_i \subset N$. Now we need to show it is normal. I think it best to demonstrate by choosing to show $\displaystyle gNg^{-1}\subset N$.
Let $\displaystyle n=n_1n_2n_3\cdots n_k \in N$ and consider $\displaystyle gng^{-1}=gn_1(g^{-1}g)n_2(g^{-1}g)n_3(g^{-1}g)\cdots(g^{-1}g) n_kg^{-1}=$ $\displaystyle (gn_1g^{-1})(gn_2g^{-1})(gn_3(g^{-1})(g\cdots g^{-1})(g n_kg^{-1})=n_1'n_2'n_3'\cdots n_k' $ by the normality of each $\displaystyle N_i$.
So unless I am really overlooking something, which I hope you guys will catch, maybe they can be shown to be contained in a normal subgroup countained in H, but not necessarily one in that original list, as alunw's counterexamples demonstrated.
*I realize that I have waved my hands about this being a subgroup even, but I know there is a theorem that if H and K are subgroups, then HK is a subgroup if and only if HK=KH. Then you can show that if $\displaystyle H\leq N_G(K)$, you do have that HK=KH and therefore HK is a subgroup. We are certainly in this case since all of these $\displaystyle N_i$ are normal, so $\displaystyle N_G(K)=G$. Perhaps a better proof would be pursued using induction upon further reflection, but it seems a pretty straightforward adjustment, and I will forgo writing it up until I see comments from others. (My hesitancy to use induction from the beginning was in the hopes that the proof could be extended to include the non finite case)
This is a reply to both of you. I'm going to come up with a conjecture of my own at the end.
First a quibble about your notation aman_cc:
Usually the notation N1*N2 would denote the direct product of the two groups N1,N2. That certainly isn't a subgroup of H, or G either, though under certain conditions it might be isomorphic to one: it is a standard theorem that the direct product of two normal subgroups N1,N2 with trivial intersection is isomorphic to the subgroup N1N2.
I assume you mean the set of elements of G that can be expressed as a product n1n2 where n1 is in N1 and n2 in N2 that is usually denoted by N1N2. Note that for general subgroups of G that is a subset of G but not a subgroup at all.
I shall prove that this is a subgroup if N2 is normal.
Consider the product of two elements n1n2 m1m2 . By the normality of N2 the element m1^-1n2m1 is in N2 and so for some p2 in N2 we have n2m1=m1p2, and hence n1n2m1m2=n1m1p2m2 = q1q2 for some q1 in N1 and q2 in N2, hence the multiplication is closed.
And m1m2^-1 = m2^-1m1^-1=m1^-1p2 for some p2 in N2 by almost the same argument.
Clearly we can argue similarly if N1 is normal. So we have shown that if G1G2 are subgroups of G then G1G2 is a subgroup if either G1 or G2 is normal.
Gamma's argument shows that G1G2 is normal if both G1 and G2 are normal.
So in fact you are right, if you have a collection of subgroups of H which are normal in G there is some subgroup of H that contains all of them and is normal in G.
My conjecture is that if N is normal subgroup of G and it contains the centre as a proper subgroup and N < H < G then H is normal in G. I'll be more than happy if someone can shoot that down in flames, (and I'm going to see if I can either prove it or come up with a counter-exaxmple) but I think it is true for the groups I am familiar with.
Well offhand its gotta be a little stricter than that because {1} is normal and in any simple group every cyclic group and other subgroup would be a counter example, I definitely do not think it is true, and Ill work on a more legitimate counter example.
Now that looks more reasonable
I really shouldn't call it a conjecture, as it's purely speculation. I was thinking about wallpaper groups and actually it's quite possibly not true even for them.
Your {1} counter-example occurred to me as soon as I'd pressed the Submit button on my last post. I think the revised version probably still is wrong but I'm not thinking straight yet.
I think it is a theorem that if a group has a subgroup of finite index then there is a subgroup of that which still has finite index and is normal. It is certainly a theorem that
if you can find a homomorphism from an infinite group onto a finite group then the kernel (obviously a normal subgroup) has finite index.
How about general linear group over the reals under times. So the center is scalar matrices $\displaystyle rI_n, r\in \mathbb{R}-\{0\}$. The subset of diagonal matrices would be a normal subgroup containing the center properly. What about upper triangular matrices being your H.
Then for instance
$\displaystyle \begin{pmatrix}0 &1\\1&0 \end{pmatrix}\begin{pmatrix}1 &1\\0&1 \end{pmatrix}\begin{pmatrix}0 &1\\1&0 \end{pmatrix}=\begin{pmatrix}1 &0\\1&1 \end{pmatrix}$
I think any finite simple group would provide a counter example to that, unless of course the trivial subgroup counts.
I need to hit the sack, its going on 5am here! It has been a pleasure discussing these sorts of things with you, look forward to many more in the future. It is these kinds of discussions where I truly feel the passion for mathematics (and learn the most)!
bah, you are right, I was too busy making sure it was actually a subgroup, didn't even think about it not being normal. Gotta be a counter example somewhere, otherwise I would think I would know this theorem. thought I was gonna be able to sleep well tonight I'll keep thinking
In other news, I did actually find an exercise in my textbook (Dummit and Foote 3.1 problem 23) "Prove that the join of any nonempty collection of normal subgroups of a group is a normal subgroup." So our proof is definitely correct.