A simple question about isomorphisms

[bmiceli21]

A "simple" question about isomorphisms

I was pondering the following question: Suppose we are given two finite groups, G and H, of order n with the property that for all m in {1,2,3,...,n}, if G has a(m) elements of order m, then so does H. Are G and H isomorphic?

Now, this seems like a natural question to ask, yet I cannot find, anywhere I look, anything about this being true or false. I would love if somebody could point me in the direction of a proof to the affirmative, or give a nice counterexample.

Thanks, in advance.

 

[The Bob]

I hope someone will correct me but I think the important question here is do the groups form a bijection? Answer that, and you more or less have the solution.

I hope I've understood the question.

The Bob

 

[bmiceli21]

The two groups have the same size, so there are definitely bijections between them, n! of them to be exact. However, the question is asking whether anyone of those is guaranteed to preserve the operation.

I know that this is really not a simple question, but I just think it seemed like a natural thing to ask, and so I figured that somebody would be able to point me in the right direction because I figured that somebody had thought about this before and the result would be easy to find.

 

[The Bob]

so there are definitely bijections between them, n! of them to be exact.

Do you mean n! (as in n factorial) bijections? Either way, two groups are only isomophic if there is a bijection between them. It is not necessarily true that two groups of the same size form a bijection.

I, again, hope to be corrected but I simply cannot fathom what you're asking.

I hope you don't think I'm being pernickety (or worse: rude).

The Bob

 

[HallsofIvy]

Do you mean n! (as in n factorial) bijections? Either way, two groups are only isomophic if there is a bijection between them. It is not necessarily true that two groups of the same size form a bijection.

I, again, hope to be corrected but I simply cannot fathom what you're asking.

I hope you don't think I'm being pernickety (or worse: rude).

The Bob

No, I don't think you are being persnickety or rude but you do seem to be using a strange definition of "bijection". A bijection is any function from one set to another that is both "one-to-one" and "onto" (both a surjection and an injection). If two sets have the same cardinality then there exist a bijection between them (that is basically the definition of "bijection") whether the sets are groups or not. An isomorphism from one group to another is a bijection that "preserves" the operation: f(a*b)= f(a)*f(b). Two groups of the same size do have a bijection between them but not necessarily an isomorophism.

 

[The Bob]

An isomorphism from one group to another is a bijection that "preserves" the operation: f(a*b)= f(a)*f(b). Two groups of the same size do have a bijection between them but not necessarily an isomorophism.

Thanks for this. So there is a bijection between the product of the cyclic groups Z₂ X Z₄ and Z₈? Is this defined (e.g. f(u) = v) or is this definition arbitrary (e.g. we design the map)? Because I know this is not an isomorphism and this is perhaps what I'm making an allusion to.

Thanks again.

The Bob

 

[HallsofIvy]

Thanks for this. So there is a bijection between the product of the cyclic groups Z₂ X Z₄ and Z₈? Is this defined (e.g. f(u) = v) or is this definition arbitrary (e.g. we design the map)? Because I know this is not an isomorphism and this is perhaps what I'm making an allusion to.

Thanks again.

The Bob

Sure there is. Just list the members of each group. For example, we can label the members of Z₂ in the obvious way: 0, 1; and then list the members of Z₄: 0, 1, 2, 3, so that the members of Z₂XZ₄ are
{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (1, 1), (1, 2), (1, 3)}. 8 members.

Now list the members of Z₈: 0, 1, 2, 3, 4, 5, 6, 7 . Also 8 members.
Then f(x) can be defined by
f(0)=(0, 0) f(1)= (0, 1), f(2)= (0, 2), f(3)= (0, 3), f(4)= (1, 0), f(5)= (1, 1), f(6)= (1, 2), and f(7)= (1, 3). That is a bijection from Z₈ to Z₂X Z₄.

It is not an isomorphism because I have paid no attention at all to the operation and, obviously, changing the order of either or both sets would give different bijections. "Injection", "surjection", and "bijection" relate to sets. "Homomorphism" and "isomorphism" relate to algebraic structures which are sets together with some operation. An isomorphism can be defined as a function on an algebraic structure that is both a bijection and a homomorphism.

 

[The Bob]

Then f(x) can be defined by
f(0)=(0, 0) f(1)= (0, 1), f(2)= (0, 2), f(3)= (0, 3), f(4)= (1, 0), f(5)= (1, 1), f(6)= (1, 2), and f(7)= (1, 3). That is a bijection from Z₈ to Z₂X Z₄.

[...] "Injection", "surjection", and "bijection" relate to sets. "Homomorphism" and "isomorphism" relate to algebraic structures which are sets together with some operation. An isomorphism can be defined as a function on an algebraic structure that is both a bijection and a homomorphism.

Brilliant. Thank you. This has cleared up a very basic mistake that not a single lecturer has ever picked up on. I really appreciate it.

So back to bmiceli21's original question, we need to show that it's a homomorphism? I think I might watch how this question unfolds from the sidelines.

Thanks again,

The Bob

 

[CompuChip]

And it is kind of obvious, that you need to map each element to an element with the same order, and that this is possible if a(m) is the same for both groups. The question is if we can always do that in such a way as to preserve the operation.

 

[The Bob]

Thanks CompuChip. I now understand the question. Experience would tell me that the operation would be preserved but I cannot provide a proof. Probably would be easier to find a counterexample.

The Bob

 

[bmiceli21]

A SOLUTION! (Re: A "simple" question about isomorphisms)

The statement is false.

Consider the groups G=Z3+Z3+Z3 (external direct product) and H, which consists of 3x3 upper-triangular matrices with 1's on the diagonal and the super-diagonal entries from Z3. Here, the operation in G is component-wise addition modulo 3 and in H we use regular matrix multiplication and then reduce the elements modulo 3.

Then |G| = |H| = 27, and each of G and H contain 26 nonidentity elements of order 3, however, they cannot isomorphic because G is Abelian and H is not.

(This couterexample is due to a colleague of mine named Jason Bell at Simon Fraser. I do not know where he found it.)

 

[The Bob]

Hey bmiceli21,

Glad you've got a solution and thanks for posting it. Apologies for the misunderstandings on my part and well done to your colleague.

The Bob