# Finding Subgroups

• Jul 25th 2010, 03:01 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:

Any help is very appreciated.
• Jul 25th 2010, 04:57 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:

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

Tonio
• Jul 25th 2010, 10:36 AM
HallsofIvy
And, of course, to do that you just have to look down the diagonal in your table.
• Jul 28th 2010, 03:38 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?
• Jul 28th 2010, 03:42 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