Math Forum :: View topic – Three problems on group

I am sorry I was wrong for question 2. Thanks very much for pointing it out.

Suppose . By considering the prime factorization of , we get , where . So is the desired element. Your answer is wrong since is not the l.c.m. of .

For question 3, after conclude that , you can consider another element . As is finite, we can proceed the same argument finitely many times and conclude. I am sorry I was gong too fast. This settles your doubt about this question.

Let me explain what we will get if . Since has at least elements in , we conclude .( I will prove the middle equality in the next paragraph). But by question 2, has an element of order , which shows it is cyclic.

To see the equality, we generally consider two subgroups and in . (Here may not be abelian hence may not be a subgroup)Write , where and the union is disjoint. So . To find , consider where .( maps cosets to cosets). Clearly is onto. To see it is injective, note that if , , implying that . Now we have shown the number of cosets equals that of , for running through . As , we have and complete the proof.

The above paragraph asnwers the second question mark made in my first solution. I am sorry that I made a mistake, since I am not so familiar with algebra as some of the other visitors do. I hope it is clear enough for you now._________________Sometimes Truth is meanlingless;What means is how you believe in.


Last edited by Milton on Thu Sep 02, 2004 2:33 pm; edited 2 times in total