In which we wonder what a non-integer lecture is anyway.
OK, of course there isn’t a lecture 8.5. But I promised that I’d post with questions for you to consider before lecture 9, so here is that post.
Preparation for Lecture 9
Lagrange’s theorem says that the order of a subgroup divides the order of the group, and a consequence of that is that the order of an element divides the order of the group. I was careful to stress that the converse is not true: if divides the order of the group, there might not be an element of order (although of course there might be). Sometimes, though, the world is a very nice place. Can you show that if a group has even order then it contains an element of order 2?
We have seen that one consequence of Lagrange’s theorem is that every group of prime order is cyclic (Corollary 37). So up to isomorphism there is only one group of order 7, for example (that is, every group of order 7 is isomorphic to the cyclic group ).
What are the possible groups of order 4? Or 6? Or 8? What are the groups of order , where is prime?
And here’s a question that we shan’t address in lecture 9 but that will be really important soon after, so I thought I’d mention it now so that you have plenty of time to think about it.
We’ve come across (left) cosets of a subgroup in a group . They’re the translates where . Intriguing question: can we impose a (natural) group structure on the set of left cosets of in ? In order to do that, we’d need to define a binary operation on : we’d want a way to take two elements and and from them obtain an element of . It seems to me that there are some very plausible guesses for what that operation might be … but then there’s some checking to be done to see whether it really does give a legitimate binary operation and whether the group axioms are satisfied.
I think that I’m going to leave that question as it is for now, but maybe I’ll come back to it in my post on lecture 9 and tell you then what the natural binary operation is and leave you to ponder whether it’s legitimate.