Is G isomorphic to a subgroup of G/M\times G/N?

In summary, the conversation discusses the concept of isomorphism, quotient groups, and homomorphisms in the context of a problem involving normal subgroups of a group G. The goal is to prove that G is isomorphic to a subgroup of the product G/M x G/N, where M and N are normal subgroups of G with only the identity element in common. Ideas are discussed, including constructing an injective homomorphism and using the properties of abelian groups.
  • #1
PingPong
62
0

Homework Statement


Let M and N be normal subgroups of G, and suppose that the identity is the only element in both M and N. Prove that G is isomorphic to a subgroup of the product [itex]G/M\times G/N[/itex]

Homework Equations


Up until now, we've dealt with isomorphism, homomorphisms, automorphisms, Lagrange's Theorem, and other bits and pieces of theorems.

The Attempt at a Solution



I have no idea how to start this! I've looked over my notes and couldn't find anything obvious. I know that |M| divides |G| and the same goes for |N|. Does the fact that M and N only have {e} as a common element mean that |N| and |M| are relatively prime?

Also, can anybody explain exactly what, say, G/M is doing? My notes mention that it means "the cosets of M in G", but I'm not sure how to deal with it. Is there an official name for it so I can look up other information on it? Thanks!
 
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  • #2
G/M is called a "quotient group". It's the group you get once you basically set M to zero.

If M and N intersect in only {e}, then that doesn't mean |M| and |N| are coprime. For example in the group C_2 x C_2 = {e, a, b, c} (the Klein four group), {e, a} and {e, b} are subgroups of order 2 that intersect only in {e}.

How about you try to construct an injective homomorphism into G/M x G/N?
 
  • #3
My understanding of quotient groups is still fuzzy... we don't have a textbook other than a supplemental online textbook, and from what I've gathered from it, G/N is only a quotient group if N is normal in G (is this right?). So what if N is not normal in G... what exactly is this object and how can I use it? I think if I understand this, it may help...

I can also see that the product MN is an abelian subgroup of G. My instructor also stated (without proof) the following two properties:

1) Every finite abelian group is a direct product of cyclic groups of prime power order
2) The prime powers that occur are uniquely determined, up to reordering

Can I use these two facts at all (especially since MN is an abelian subgroup of G)?

As another idea, I tried going back over my notes and came up with something else of interest. We showed in class that for a homomorphism [itex]\phi:G\rightarrow H[/itex], the image of G is a subgroup of H. If the homomorphism if injective, then it is certainly surjective onto its own image, which means that it is an isomorphism. Is this what you were suggesting, morphism?

If that's the case, then all I need are surjective homomorphisms from G to G/M and G/N, right? But I can't really do that until I understand quotient groups a bit more... :( I'll keep at it! Thanks for your help morphism!

EDIT: Whoops, I meant I need injective homomorphisms. Sorry!
 
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  • #4
PingPong said:
My understanding of quotient groups is still fuzzy... we don't have a textbook other than a supplemental online textbook, and from what I've gathered from it, G/N is only a quotient group if N is normal in G (is this right?). So what if N is not normal in G... what exactly is this object and how can I use it? I think if I understand this, it may help...
Let's introduce an equivalence relation on G. We'll say that g~h iff gh-1 is in N. This is motivated by our desire to make N=e. Consider the set G/~ of equivalence classes of G under ~. Let's use the notation gN to denote the equivalence class to which g belongs. Now our goal is to turn G/~ into a group. The natural operation that comes to mind is gN * hN = (gh)N -- but we need to ensure that this is well-defined. That is to say, if gN=g'N and hN=h'N (so g~g' and h~h'), then we must have (gh)N=(g'h')N (so gh~g'h'). You can check that the N being normal in G is exactly what we need for this to work. That G/~ is a group is easy to verify (for instance the identity element is eN and the inverse of gN is g-1N), and we call it the quotient group of G by N, which we denote by G/N.

As another idea, I tried going back over my notes and came up with something else of interest. We showed in class that for a homomorphism [itex]\phi:G\rightarrow H[/itex], the image of G is a subgroup of H. If the homomorphism if injective, then it is certainly surjective onto its own image, which means that it is an isomorphism. Is this what you were suggesting, morphism?
Yup, that's what I'm suggesting. Just set up the most obvious map from G to G/M x G/N you can think of (remember, the elements of the latter are ordered pairs (gN, gM)), and check that it's an injective homomorphism.

The abelian group stuff isn't going to be very helpful here at all. (And MN doesn't have to be abelian. For example, S_4 contains C_2 x C_2 and A_4 as normal subgroups, and their product, A_4, is nonabelian.)
 
  • #5
Okay, so please correct me if I'm wrong about this then:

Fix any n in N and any m in M and make a homomorphism [itex]\phi:G\rightarrow (G/N\times G/M)[/itex] defined by [itex]\phi(g)=(g n , g m)[/itex]. Can I say for sure that this is injective? Certainly, each component is not (necessarily) injective, that is, f(g)=gn is not injective into gN, because, unless N={e}, |G/N| < |G|, right? Is this where the bit about M and N having only the identity in common comes in?

Sorry if I seem a bit dense. Your help is very much appreciated, morphism!
 
  • #6
PingPong said:
Okay, so please correct me if I'm wrong about this then:

Fix any n in N and any m in M and make a homomorphism [itex]\phi:G\rightarrow (G/N\times G/M)[/itex] defined by [itex]\phi(g)=(g n , g m)[/itex]. Can I say for sure that this is injective? Certainly, each component is not (necessarily) injective, that is, f(g)=gn is not injective into gN, because, unless N={e}, |G/N| < |G|, right? Is this where the bit about M and N having only the identity in common comes in?

Sorry if I seem a bit dense. Your help is very much appreciated, morphism!

[itex]\phi(g)=(g n , g m)[/itex] doesn't make sense, elements of [itex]G/N\times G/M[/itex] are ordered pairs of cosets, i think you want [itex]\phi(g)=(gN, gM)[/itex]

use the definition of injective, and explain where you get stuck, remember g_1N = g_2N iff (g_1)^-1g_2 is in N, likewise for M.
 
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1. What is a subgroup of a direct product?

A subgroup of a direct product is a subset of the direct product that also forms a group under the same operation. It contains elements that are formed by combining elements from each individual group in the direct product.

2. How do you determine if a subset is a subgroup of a direct product?

To determine if a subset is a subgroup of a direct product, you need to check if it satisfies the three conditions of a subgroup: closure, associativity, and inverse elements. The subset must be closed under the operation, associative, and each element must have an inverse within the subset.

3. Can there be multiple subgroups in a direct product?

Yes, there can be multiple subgroups in a direct product. In fact, the direct product of two groups can have infinitely many subgroups.

4. What are the benefits of studying subgroups of a direct product?

Studying subgroups of a direct product can help in understanding the structure and properties of the direct product, as well as the individual groups that make up the direct product. It can also aid in solving problems and proving theorems in abstract algebra.

5. How are subgroups of a direct product related to factor groups?

Subgroups of a direct product and factor groups are closely related concepts. In fact, subgroups of a direct product are the building blocks for factor groups. A factor group is formed by taking a subgroup of the direct product and dividing it out, or factoring it, from the original group.

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