linzedun Posted November 23, 2004 Share Posted November 23, 2004 Question: Show that S4 has a unique subgroup of order 12. So far I know that A4 is a subgroup of order 12. So I was trying to prove by contradiction that there was another subgroup of order 12, say G. So the (|G||A4|/the intersection of G and A4) = 24. Then I need to show something to the effect that the intersection of G and A4 has to be a subgroup of order 6 in A4 and that that is a contradiction. This is as far as I got and my teacher says I am on the right track, but I have no ideas of where to go next, help please! Link to comment Share on other sites More sharing options...
matt grime Posted November 24, 2004 Share Posted November 24, 2004 Hint G is also normal. Link to comment Share on other sites More sharing options...
linzedun Posted November 24, 2004 Author Share Posted November 24, 2004 So I can prove that A4 has no subgroup of order 6 and then I have it proved right? Link to comment Share on other sites More sharing options...
matt grime Posted November 25, 2004 Share Posted November 25, 2004 No, not as stands. Proving A4 has no subgroup of order six isn't going to help unless you have provem that the assumption G exists and isn't A4 implies that it must possess such. You don't appear to have done that. Link to comment Share on other sites More sharing options...
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