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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.

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It looks like $\mathbb{Z}_4$ to me. Why do you say you don't think it is? – Billy Aug 25 '11 at 2:31
@Billy: It is not cyclic. More precisely we have $a^2=e$ and $b^2=a$. – user532 Aug 25 '11 at 2:32
It is cyclic. It's generated by b. Note that a = b^2, ba = b^3. :) – Billy Aug 25 '11 at 2:32
@Billy: Accepting answers isn't just about reputation; it's also about marking the question as answered so people don't unnecessarily go back to it. If you feel you don't deserve reputation for the answer, you can always make it community wiki (by ticking the checkbox under the answer text area). – joriki Aug 25 '11 at 2:54
The first two words of your question are a bit of a give-away..."The group..." – user1729 Aug 25 '11 at 10:24
up vote 4 down vote accepted

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 $b$.

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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 $Z_4$.

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Your answer is implicitly assuming that it is a group and proving that it must be cyclic! Your underlying idea is correct, but if, for example, 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? – user1729 Nov 12 '14 at 12:50

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