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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?2011-08-25
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    @Billy: It is not cyclic. More precisely we have $a^2=e$ and $b^2=a$.2011-08-25
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    It is cyclic. It's generated by b. Note that a = b^2, ba = b^3. :)2011-08-25
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    Usually, when we write a table, the entry in the $i$th row and the $j$th column is $a_ia_j$; i.e., the product of the label of the row times the label. But you are making the (2,3) entry the product or the label of the 3rd column, $b$, by the label of the second row, $a$. Just so you know.2011-08-25
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    @Billy: Oh.. That totally skipped me.. Thanks :) I guess this is what learning is all about :) ! Many thanks :) Please post it as answer so I can accept it.2011-08-25
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    @Arturo Magidin: Thanks for the note, I didn't know this convention :)2011-08-25
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    @user532: Don't worry about the reputation, but thanks anyway!2011-08-25
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    @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).2011-08-25
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    @joriki: Good point. Okay, I'll submit it as an answer.2011-08-25
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    The first two words of your question are a bit of a give-away..."The group..."2011-08-25

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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?2014-11-12