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This question already has an answer here:

Assume $G = \{1, a,b,c\}$ is a group of order $4$ with identity $1.$ Assume also that $G$ has no elements of order $4$. Show that there is a unique group table for $G$. Also show that $G$ is abelian.

If $G$ is abelian, then the group table matrix must be symmetric. How can I introduce a binary function and show it? I am new in this field, so I am not so familiar. I have proved many other exercises, but it is a little tough (for me).

Can you please help?

Edit: I know every element has order $\leq 3$ , but I do not understand how I will proceed.

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marked as duplicate by Somos, Alex Provost, Eevee Trainer, Lord Shark the Unknown abstract-algebra Mar 23 at 5:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The order of the elements must divide the order of the group. Since $|G|=4$ and there's no elements of order $4$ all elements must have order $2$ or order $1$ (clearly the latter must be the identity).

The element $ab$ must be in the group. $ab\not = e$ because $a^2=e$ and this would imply $a=b$. Also $ab\not = a$ because then $b=e$ and $ab\not = b$ because then $a=e$. In other words $ab=c$.

Symmetric argument will also give that $ba=c$. Similarly $ac=ca=b$ and $bc=cb=a$.

We completely calculated all the products in this group.

$$\left[\begin{array}{c|cccc}* & \textbf{1} & \textbf{a} & \textbf{b} & \textbf{c}\\ \hline \textbf{1} & 1 &a &b&c \\ \textbf{a} &a&1&c&b \\ \textbf{b} &b&c&1&a\\ \textbf{c}&c&b&a&1\end{array}\right]$$

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    $\begingroup$ This is a nice answer. I took the liberty of formatting your multiplication table to separate the row and column headings. $\endgroup$ – Théophile Mar 22 at 19:21
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    $\begingroup$ @Théophile I was wondering how to do that. Thank you very much! $\endgroup$ – Yanko Mar 22 at 19:22
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    $\begingroup$ You're welcome! There's an excellent MathJax reference that shows how to do all sorts of other things. $\endgroup$ – Théophile Mar 22 at 19:35
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Just use juxtaposition for the binary operation.

Further, here is a . . .

Hint: If $a^2=b$, then $ab=a^3\neq 1$ by Lagrange's Theorem.

The group is (isomorphic to) $\Bbb Z_2\times \Bbb Z_2$.

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    $\begingroup$ @ShamimAkhtar That's magnanimous of you. :) It's better to accept an answer, though, so that the question no longer appears in the list of "unanswered questions". $\endgroup$ – Théophile Mar 22 at 19:33
  • $\begingroup$ Ok then i was thinking they will feel bad $\endgroup$ – user655794 Mar 22 at 19:34
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    $\begingroup$ Théophile is right, @ShamimAkhtar; thank you nonetheless :) $\endgroup$ – Shaun Mar 22 at 19:40