# Showing there is a unique group table for $\{1, a,b,c\}$ such that there is no element of order $4$. [duplicate]

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).

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

## marked as duplicate by Somos, Alex Provost, Eevee Trainer, Lord Shark the Unknown abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 23 at 5:27

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.

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

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

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) $$\Bbb Z_2\times \Bbb Z_2$$.

• @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". – Théophile Mar 22 at 19:33
• Ok then i was thinking they will feel bad – user655794 Mar 22 at 19:34
• Théophile is right, @ShamimAkhtar; thank you nonetheless :) – Shaun Mar 22 at 19:40