# Math Help - Generalized proof of isomorphism: Once more unto the breach!

1. ## Generalized proof of isomorphism: Once more unto the breach!

I'm going to have to post the full form of this question as I doubt the terminology is universal. So here it goes!

Let R be the set of all polynomials with integer coefficients in the independent variables x_1, x_2, x_3, x_4, ie. the members of R are finite sums of elements of the form $ax_1^{r_1}x_2^{r_2}x_3^{r_3}x_4^{r_4}$ where a is any integer and r_1, ..., r_4 are nonnegative integers. Each $\sigma \in S_4$ gives a permutation of {x_1, ..., x_4} by defining $\sigma \cdot x_i = x_{\sigma (i)}$. This may be extended to a map from R to R by defining $\sigma \cdot p( x_1, x_2, x_3, x_4 ) = p( x_{\sigma (1)}, x_{\sigma (2)}, x_{\sigma (3)}, x_{\sigma (4)}$ for all $p( x_1, x_2, x_3, x_4 ) \in R$.

Exhibit all permutations in S_4 that stabilize the element $x_1 x_2 + x_3 x_4$ and prove that they form a subgroup isomorphic to the dihedral group D_8.
Okay, after that mouthful I am going to sketch the method I used to answer the question. I found the stabilizer, then calculated the order of each element. I did this also for each element of D_8, then constructed a bijective function between the two groups. It was a bit tedious, but I then verified that this was also an homomorphism. So my bijection is an isomorphism between the normalizer and D_8.

My question is: Is it possible to find the isomorphism more generally, rather than constructing the bijection?

-Dan

2. ## Re: Generalized proof of isomorphism: Once more unto the breach!

Dan,
Presumably you found that the stabilizer H contains x = (1 3 2 4) and y = (1 2). Easily x4 = y2 = (1) and y-1xy = x-1. So H contains < x, y : x4 = y2 = (1) and y-1xy = x-1>. This last description is a presentation of a group via generators and relations. In order to understand this idea, you have to look at "free" groups. It turns out that any two groups with the same presentation are isomorphic. The dihedral group of order 8 is D8 = < r, s : r4 = s2 = (1) and s-1rs = s-1>. So D8 is isomorphic to a subgroup of H. Since this subgroup has index 3 in S4 and H is unequal to S4, H is equal to this subgroup.

So in sum, the way to prove generally two groups are isomorphic is to show that they have the same presentation.

3. ## Re: Generalized proof of isomorphism: Once more unto the breach!

Originally Posted by johng
D8 = < r, s : r4 = s2 = (1) and s-1rs = s-1>.
Possibly too obvious to mention, but in case it confuses someone, johng meant $D_8 = \langle r,s: r^4=s^2=(1)\text{ and }s^{-1}rs = r^{-1} \rangle$ (the final $s$ above should be an $r$).

4. ## Re: Generalized proof of isomorphism: Once more unto the breach!

Here is another way: suppose you have already established that the stabilizer has order 8. You then try to narrow down the isomorphism class by asking the following questions:

1) Is the stablizer abelian?

1a) If so, is it cyclic?

1a.1) If not cyclic (but abelian), does it have any element of order 4? (Equivalently, are there 7 elements of order 2, or just 3?)

1b) If non-abelian:

2) Does it have 5 elements of order 2, or just 1? (this amounts to deciding between Q8 and D8).

In general, looking at elements of order 2 often reveals a great deal about which group we might have (this is especially true when the order of our group is a power of 2, the "worst case scenario"). Looking at centralizers of elements of order 2 can also be very useful.

This approach, of course, fails miserably when we are looking at finite groups of larger order (like, for example, 210, for which there are over 49 BILLION isomorphism classes).

I'd like to point out that the converse of what johng posted is not true: Non-isomorphic presentations may in fact describe the same group...in fact, it is generally regarded as a VERY difficult question, given two presentations, to decide if they in fact describe the same group. So comparing presentations is not, in general, the BEST way to determine the isomorphism class of a group (although it can be used to show isomorphism for certain well-known groups).

5. ## Re: Generalized proof of isomorphism: Once more unto the breach!

Thanks all. Presentations were exactly the thing I was looking for.

-Dan