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--R. Kersey | --R. Kersey | ||
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| + | They are equal because they describe the same set of elements, not because they have the same order. | ||
| + | :--[[User:Narupley|Nick Rupley]] 12:55, 5 February 2009 (UTC) | ||
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| + | --J. Korb | ||
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| + | Just wondering if anyone found a better way of doing this problem than writing out all the subgroups and ignoring the ones that are the same. It seems like there should be a cleaner way of doing this. | ||
Latest revision as of 11:43, 5 February 2009
List the cyclic subgrous of U(30).
To do this, I listed each possible subgroup of U(30) and then tested whether they are cyclic. A problem I ran into, though, is that <13> appears to be cyclic and <23> does, but the answer in the back of the book does not include them.
Has anyone else encountered this problem or know the solution? I will try to ask during office hours....
--A. Cadwallader
I think they aren't in the back of the book because <13> is the same as <7> and <23> is the same as <17>.
--S. Rosenberger
Is <13> = <7> and <23> = 17 because they have the same elements or the same order. What I'm asking is if cyclic groups of the same order equal to each other.
--R. Kersey
They are equal because they describe the same set of elements, not because they have the same order.
- --Nick Rupley 12:55, 5 February 2009 (UTC)
--J. Korb
Just wondering if anyone found a better way of doing this problem than writing out all the subgroups and ignoring the ones that are the same. It seems like there should be a cleaner way of doing this.
