# Group isomorphism and Polynomial ring modulo ideal

1. Aug 12, 2007

### X-il3

Hi everyone.

I have two questions that I hope you can help me with.

First when trying to show isomorphism between groups is it enough to show that the order of each element within the group is the same in the other group? For example the groups (Z/14Z)* and (Z/9Z)*. They are both of order 6 and both have
1 element of order 1
1 element of order 2
2 elements of order 3
2 elements of order 6
And does the binary operator have to be the same in both groups when doing group isomorphism?

And the second question(actually the third one). I am trying to write the multiplication table for the following ring
Z_2[X]/(x^2 + x +). I know that this ring has 4 elements. Is it correct that the elements are the following
1
x + 1
x^2 + 1
x^2 + x + 1
?

Hope you can help me with this,

X-il3

2. Aug 12, 2007

### matt grime

No.

What do you mean by 'same'? If f is an isomorphism then f(xy)=f(x)f(y), if that's what you mean.

Is there a missing 1 there?

Where is 0? Why haven't you written down x or x^x+x? (You have, by the way, but you have made a strange choice of representatives of the elements, which is why I ask, since it implies you've not really understood what you're doing.

3. Aug 12, 2007

### X-il3

Yes there was 1 missing there.

I thought the identity element in multiplication is 1 and therefore no 0. How would you represent the elements in this ring? More like this
0
x
x^2
x^2 + x
? Leaving the 1 out from all the elements?

X-il3

4. Aug 12, 2007

### matt grime

Now where's 1 gone? (Again, it is still there in disguise which makes more confused as toy what you're trying to do).

5. Aug 12, 2007

### X-il3

Ok here is one more try. Hope I get it right this time .
Since we are working in Z_2 then a = -a where a is an element in Z_2.

We know that X^2 + x + 1 = 0 = [0]. Moving numbers around now we then get the remaining 3 elements in the ring.
x^2 + x = 1 = [1]
x^2 + 1 = x = [x]
x + 1 = x^2 = [x^2]

X-il3