I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them here). The lectures come complete with notes and homework problems. Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I'm doing them correctly. So I've decided to try to crowd source the grading of my solutions. I'll post individual questions as I finish them and wait for comments. I would like to get critiques of not just my reasoning but the style of my write ups as well. Also, any alternative approaches to the problem would be welcome. I've looked for forums dedicated to these kinds of OCW courses but have not been able to find any. This surprises me. It seems like, with the advent of these free online educational resources, an online meeting place for those who take advantage of them would be a natural offshoot. So this might be a bit of an experiment. Unless I'm the 873rd person to post something like this here. If that's the case, sorry for being so long winded.
Anyway, on with the question. This one is assigned in the 4th lecture.
Let $V$ denote the Klein 4-group. Show that $\operatorname{Aut}(V)$ is isomorphic to $S_3$.
(I've moved my solution below to keep this question from showing up in the unanswered list.)
Thanks…
Best Answer
This is my answer. Additional comments are always welcome.
Let $f:V\rightarrow V$ be a permutation that fixes the identity element. By definition, this map is bijective. $V$ has the property that the product of any 2 distinct non-identity elements is the 3rd non-identity element. This implies $f(vv^\prime)=f(v)\cdot f(v^\prime)$ for all $v,v^\prime \in V$ with $v,v^\prime\neq e$. For products involving the identity element, $f(ev)=f(e)\cdot f(v)=e\cdot f(v)=f(v)$ for all $v\in V$. Also, since all non-identity elements of $V$ have order 2, $f(v^2)=f(vv)=f(v)\cdot f(v)=e$ for all $v\in V$. Hence, any permutation of the elements of $V$ that fixes the identity element is an automorphism of $V$. Since these permutations exhaust all possible automorphisms of $V$, they are the elements of Aut($V$).
Since the $3!=6$ elements of Aut($V$) represent all permutations of 3 objects, Aut($V$) is isomorphic to S$_3$.