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.)


  • 6
    $\begingroup$ Excellent. I like the general idea of "crowd sourcing" your solutions. By the way, another approach is to realize that the set of automorphisms is surely a subset of $S_3$, and so it suffices to find an automorphism of order 2 and one of order 3. $\endgroup$ – user641 Jan 11 '12 at 21:07
  • $\begingroup$ Thanks Steve. If we can state from the beginning that Aut(V) is a subset of S$_3$, would it then be enough to say that the order of Aut(V) is 6 to prove it's isomorphic to S$_3$ without having to find the automorphisms? $\endgroup$ – jobrien929 Jan 11 '12 at 22:34
  • $\begingroup$ @jobrien929Hi, I'm going through the same process. I am, however, using the 2nd edition. Do you have any suggestions as to how know what are the corresponding problems. Also, while I'm not very good at it, so I don't know what contribution I could make, if you set up a website or other mechanism, I would really appreciate being informed. Regards. $\endgroup$ – user12802 Jan 11 '12 at 23:56
  • $\begingroup$ @Andrew Nice to know I'm not doing this alone. I can't be much help with figuring out which problems you should work on in the 2nd edition. I can tell you that a few minutes with google and the keywords Artin, Algebra, and pdf would probably render the problem moot. I won't be setting up a website or anything but I will be posting my work here so if you look up my postings you'll be seeing new problems appear. I'm more than happy to discuss any of the assignments in the comments. It's the whole reason I'm taking the time to post them. Thanks. $\endgroup$ – jobrien929 Jan 12 '12 at 4:29
  • 1
    $\begingroup$ @Andrew: Every automorphism of $V$ will fix the identity, and will act as a permutation (injectively and surjectively) on the non-identity elements, of which there are $3$. Thus the automorphism group can be treated as a subgroup of $S_3$. Now consider what Lagrange's theorem says about the order of elements, and the order of the group that contains them. $\endgroup$ – user641 Jan 24 '12 at 21:51

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$_3$.

  • $\begingroup$ The answer depends on the background that is assumed. If you have done integers mod $n$, then one can note that $V$ can be given the structure of a $2$-dimensional space over $\mathbf{Z}_{2}$, with group homomorphisms being automatically linear maps. Now note that any two nonzero, distinct elements are a basis... $\endgroup$ – Andreas Caranti Feb 13 '13 at 8:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.