0
$\begingroup$

The group has two generators, $a,b$ with identity $e$ and multiplication table: $ \begin{array}{|l|l|l|} e & a & b & ba \\ \hline a & e & ba & b \\ \hline b & ba & a & e \\ \hline ba & b & e & a \end{array} $

I've looked up the list of groups of order 4 and this isn't $\mathbb Z_4$ nor $V_4$. So it can't be a group. But I am not able to locate where the problem is.

  • 1
    The first two words of your question are$a$bit of a give-away..."The group..."2011-08-25

2 Answers 2

4

The group is $\mathbb{Z}_4$: note that $a = b^2$ and so $ba = b^3$, meaning that the group is cyclic and generated by $b$.

0

Another interpretation may be $b^2=a$, and $a^2=e$, so that $b^4=e$ shows that [here] order of $b$ is 4. We know that whenever there exist any element of order $n$ in a group of order $n$ that group becomes cyclic. So this group is $Z_4$.

  • 0
    Your answer is implicitly assuming that it is a group and proving that it must be cyclic! Your underlying idea is correct, but i$f$, for e$x$ample, you got a 4x4 table and you spotted two elements of order two then you would first need to prove that it was a group before concluding that it was the Klein 4-group. Does that make sense?2014-11-12