Finding Subgroups

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July 25th 2010, 02:01 AM
demode

Finding Subgroups

Let G be the group $(U(15), \times_{15})$ . Find all distinct subgroup of G with exactly two elements.

Could anyone show me how to deal with this problem? So, I know that U(15)={1, 2, 4, 7, 8, 11, 13, 14} under multiplication modulo 15. I'm not sure if it helps, but I have constructed a Cayley table for G:

http://img534.imageshack.us/img534/6739/cayley.gif
Any help is very appreciated.
July 25th 2010, 03:57 AM
tonio

Quote:

Originally Posted by demode

Let G be the group $(U(15), \times_{15})$ . Find all distinct subgroup of G with exactly two elements.

Could anyone show me how to deal with this problem? So, I know that U(15)={1, 2, 4, 7, 8, 11, 13, 14} under multiplication modulo 15. I'm not sure if it helps, but I have constructed a Cayley table for G:

http://img534.imageshack.us/img534/6739/cayley.gif
Any help is very appreciated.

Of course it helps! A subgroup with two elements is of the form $\{1,a\}\,,\,\,with\,\,\,a^2=1$ , so locate all the elements with order 2 and
form then your subgroups...

Tonio
July 25th 2010, 09:36 AM
HallsofIvy

And, of course, to do that you just have to look down the diagonal in your table.
July 28th 2010, 02:38 AM
demode

Okay then it's all the 1's in the main diagonal: g={1, 4, 11, 14}.

And elements like 2 belong to order 4? Because $2 \times_{15} 2 = 4, 4 \times_{15} 2 = 8, 8 \times_{15} 2 = 1$ . Is that right?
July 28th 2010, 02:42 AM
tonio

Quote:

Originally Posted by demode

Okay then it's all the 1's in the main diagonal: g={1, 4, 11, 14}.

And elements like 2 belong to order 4? Because $2 \times_{15} 2 = 4, 4 \times_{15} 2 = 8, 8 \times_{15} 2 = 1$ . Is that right?

Right indeed.

Tonio