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That's what I did, too. I'm not sure if it was more or less work than Cayley tables, though. -Tim | That's what I did, too. I'm not sure if it was more or less work than Cayley tables, though. -Tim | ||
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| + | pages124-128 do an excellent job of describing how to solve this problem, in particular, the Cayley tables and the description of isomorphisms | ||
Latest revision as of 01:13, 25 September 2008
I found that they are not isomorphic because they are not "onto" (1 to 1). Did anyone else come to that conclusions? I used Cayley tables...they were big and it took some work. I'm sure there's a quicker way...
I also found that they are not isomorphic. However, I used the fact that two groups must preserve order to be isomorphic. So first looking at the number of elements in each group we see that:
U(20) = {1,3,7,9,11,13,17,19} and U(24) = {1,5,7,11,13,17,19,23}.
For U(20):
|1|= 1
|3|= 4
|7|= 4
|9|= 2
|11|= 2
|13|= 4
|17|= 4
|19|= 2
For U(24):
|1|= 1
|5|= 2
|7|= 2
|11|= 2
|13|= 2
|17|= 2
|19|= 2
|23|= 2
So even though both groups have the same number of elements, we can see that the order of the elements is not equivalent. Therefore, order is not preserved and the two groups cannot be isomorphic. -Jesse
That's what I did, too. I'm not sure if it was more or less work than Cayley tables, though. -Tim
pages124-128 do an excellent job of describing how to solve this problem, in particular, the Cayley tables and the description of isomorphisms
