# Isomorphic Refinement

• Mar 8th 2010, 10:59 AM
Ryaη
Isomorphic Refinement
Given {0} < {18} < {3} < $Z_{72}$ and {0} < {24} < {12} < $Z_{72}$ , would the isomorphic refinement be:

{0} < {18} < {3} < $Z_{72}$ and {0} < {24} < {12} < {4} < $Z_{72}$, respectively?

I came to that as my solution because the refinements would then have the factor groups isomorphic to $Z_{3}, Z_{6}, Z_{24}, Z_{4}, Z_{72}, and Z_{18}$.

It should be stated that I don't quite fully understand isomorphic refinements, and I took it to mean that every possible pair in one series needs to have an isomorphic match in the other series.
• Mar 8th 2010, 04:56 PM
aliceinwonderland
Quote:

Originally Posted by Ryaη
Given {0} < {18} < {3} < $Z_{72}$ and {0} < {24} < {12} < $Z_{72}$ , would the isomorphic refinement be:

{0} < {18} < {3} < $Z_{72}$ and {0} < {24} < {12} < {4} < $Z_{72}$, respectively?

I came to that as my solution because the refinements would then have the factor groups isomorphic to $Z_{3}, Z_{6}, Z_{24}, Z_{4}, Z_{72}, and Z_{18}$.

It should be stated that I don't quite fully understand isomorphic refinements, and I took it to mean that every possible pair in one series needs to have an isomorphic match in the other series.

Let
1: {0} < <18> < <3> < $\mathbb{Z}_{72}$ and
2: {0} < <24> < <12> < $\mathbb{Z}_{72}$.

The first series has quotient groups of order 4, 6, 3 and the second series has quotient groups of order 3, 2, 12. (see here for details)

Now split 4 into 2 and 2 while 12 into 6 and 2.

Then we can make the first series has a factor group of order 2,2,6,3 and the second series has a factor group 3,2,2,6. Now the resulting isomorphic refinements are:

1: {0} < <36> < <18> < <3> < $\mathbb{Z}_{72}$ and
2: {0} < <24> < <12> < <6> < $\mathbb{Z}_{72}$.
• Mar 8th 2010, 07:16 PM
Ryaη
Quote:

Originally Posted by aliceinwonderland
The first series has quotient groups of order 4, 6, 3 and the second series has quotient groups of order 3, 2, 12. (see here for details)

Even after looking at the details I'm not exactly sure how you came up with these numbers.

Once you have them, I think I understand how and why you split them though.
• Mar 8th 2010, 08:18 PM
aliceinwonderland
Quote:

Originally Posted by Ryaη
Even after looking at the details I'm not exactly sure how you came up with these numbers.

Once you have them, I think I understand how and why you split them though.

If you don't split at all, they are not isomorphic refinements because it is clear that the orders of factor groups 4, 6, 3 and 3, 2, 12 are not the same at all. Try to factor the biggest number 12 and make both series the same up to the order.
Let's say we factor 12 into 2 and 6. Then the second series become 3,2,2,6. Now it is clear how we factor a number of the first series and make the same series with the second series up to the order.
• Mar 9th 2010, 10:52 AM
Ryaη
Oh! I think I get it now! Thanks!
• Mar 9th 2010, 12:01 PM
Ryaη
Quote:

Originally Posted by aliceinwonderland
Let
1: {0} < <18> < <3> < $\mathbb{Z}_{72}$ and
2: {0} < <24> < <12> < $\mathbb{Z}_{72}$.

The first series has quotient groups of order 4, 6, 3 and the second series has quotient groups of order 3, 2, 12. (see here for details)

Now split 4 into 2 and 2 while 12 into 6 and 2.

Then we can make the first series has a factor group of order 2,2,6,3 and the second series has a factor group 3,2,2,6. Now the resulting isomorphic refinements are:

1: {0} < <36> < <18> < <3> < $\mathbb{Z}_{72}$ and
2: {0} < <24> < <12> < <6> < $\mathbb{Z}_{72}$.

So, is a very simpleton way of deducing this for the first series, that:

72/18 = 4, 18/3 = 6, 3/0 = 3; and then any other number you could come up with would just be some multiple or multiplication of these numbers?

And then when you split the 4, you have to add in 36 to keep the property mentioned above?

I feel like that could be very wrong, but it was the only way that I was able to determine that 4, 6, and 3 are quotient groups of the first series.

This all goes ditto for the second series.
• Mar 9th 2010, 01:53 PM
aliceinwonderland
Quote:

Originally Posted by Ryaη
So, is a very simpleton way of deducing this for the first series, that:

72/18 = 4, 18/3 = 6, 3/0 = 3; and then any other number you could come up with would just be some multiple or multiplication of these numbers?

And then when you split the 4, you have to add in 36 to keep the property mentioned above?

I feel like that could be very wrong, but it was the only way that I was able to determine that 4, 6, and 3 are quotient groups of the first series.

This all goes ditto for the second series.

What matters is the order of the factor groups of a "consecutive pair" as shown by the above link. The bottom line is that if the two series are isomorphic refinements, then the set of these numbers should be the same (up to the rearrangement) having the same cardinality.