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Thanks - [[User:Eraymond|Eraymond]] 00:08, 26 February 2009 (UTC) | Thanks - [[User:Eraymond|Eraymond]] 00:08, 26 February 2009 (UTC) | ||
| + | |||
| + | ---- | ||
| + | For this problem I did something similar to the previous problem. From theorem 6.5 we can say that Aut(Z_20) = U(20). From this we get:<br> | ||
| + | Aut(Z_20) = U(20) = U(5x4) = U(5) x U(4) = Z_4 x Z_2<br> | ||
| + | Now we can observe that Z_4 = {0,1,2,3} and Z_2 = {0,1} and from this we can iterate the elements of orders 2 and 4. The elements of order 2 are {(0,1), (2,1), (2,0)} and the elements of order 4 are {(1,0), (1,1), (3,0), (3,1)}.<br> | ||
| + | Emily if you want to try and do it that way then look at the post for number 34. I try and explain that theory via example on that page. If it's still unclear I can try and explain it better.<br> | ||
| + | --[[User:Jniederh|Jniederh]] 00:45, 26 February 2009 (UTC) | ||
Revision as of 20:45, 25 February 2009
I have a question about finding out the choices for lcm(|a|, |b|) = 4. On page 161 in the new book, it goes into a big explanation for a different example and I am having trouble following.
Thanks - Eraymond 00:08, 26 February 2009 (UTC)
For this problem I did something similar to the previous problem. From theorem 6.5 we can say that Aut(Z_20) = U(20). From this we get:
Aut(Z_20) = U(20) = U(5x4) = U(5) x U(4) = Z_4 x Z_2
Now we can observe that Z_4 = {0,1,2,3} and Z_2 = {0,1} and from this we can iterate the elements of orders 2 and 4. The elements of order 2 are {(0,1), (2,1), (2,0)} and the elements of order 4 are {(1,0), (1,1), (3,0), (3,1)}.
Emily if you want to try and do it that way then look at the post for number 34. I try and explain that theory via example on that page. If it's still unclear I can try and explain it better.
--Jniederh 00:45, 26 February 2009 (UTC)
