mt87 Posted October 4, 2005 Share Posted October 4, 2005 If G is an abelian group and n>1 an integer, let A={a^n such that a E G}. Prove that A is a subgroup of G. isn't the identity of A a^0 which does not fall under n>1 Link to comment Share on other sites More sharing options...
Dave Posted October 4, 2005 Share Posted October 4, 2005 You're missing something. A is defined using every element from G, so if the identity is an element in G then it's trivially in A. Link to comment Share on other sites More sharing options...
mt87 Posted October 4, 2005 Author Share Posted October 4, 2005 how does the fact that the group is albelian help? I know that every cyclic group is abelian but in this case it is the subgroup that is cyclin. Does this proof need the basic prove-the-axioms-of-a-group approach or is it more complex? Link to comment Share on other sites More sharing options...
Algebracus Posted October 6, 2005 Share Posted October 6, 2005 You just check the axioms. For the existence of an inverse, we use the fact that G is abelian: a^n * (a^-1)^n = (a * a^-1)^n = ... Link to comment Share on other sites More sharing options...
Dave Posted October 6, 2005 Share Posted October 6, 2005 Don't know whether this will help, but sometimes it saves a bit of time. If G is a group with H a subset of G, then if you can prove the condition [imath]g, h \in H \Rightarrow gh^{-1} \in H[/imath], H is a subgroup. As I said, it's saved me some time in exams, so there we go Link to comment Share on other sites More sharing options...
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