I should consider group tables obtained by renaming elements as essentially the same and then show that there are only two essentially different groups of order 4.

There seems to be so many different possible group tables for all the different binary operations - which is why I'm confused. I was thinking about using Cayley's table to show their commutativity but I'm really not too sure. Any help please!

Edit: After all of your help, I completely understand how to show that there are only two different groups of order four. Thank you. The only thing which I am still unclear of is how to note down the 'other tables' before stating that they are essentially the same as one of the other tables - meaning that there are just two different ones. It's pointless work but I think it's what the question requires.

Answer: After a bit of playing around - I've realised that there are four different tables, however, 3 tables are the same as each other, just with different values (the Klein 4 Group with 3 different generators). Hence there are two tables.

  • $\begingroup$ It might seem like there are many possibilities, but try to start writing out some of them and note that there are certain properties they must have in order to satisfy the group axioms. $\endgroup$ – Tobias Kildetoft Feb 11 '13 at 17:47
  • $\begingroup$ Re: duplicate vote. It's the same question, but the discussion in the older one is left unfinished. $\endgroup$ – user53153 Feb 11 '13 at 19:12

Hints: we are talking about groups of order 4, which narrows the possible binary relations and elements available:

  • each must contain the identity,
  • each must be associative,
  • each must be closed under inverses (if $a \in G,\;a^{-1} \in G$)
  • (and of course, closed under the group operation).

There are essentially (up to isomorphism) only two groups of order 4:

  • one of course will be the additive cyclic group $\mathbb{Z}_4$, and
  • the other will be the Klein 4-group, which is indeed abelian.

If you follow the suggestions for solving the problem, you'll find, indeed, that any possible GROUP of order 4 can be shown isomorphic to one of the two groups mentioned simply by a renaming elements.

Yes: use of the Cayley table will be of great importance:

  • for a group: no element can appear twice in any column,
  • no element can appear twice in any row.

You'll find that the only ways of completing a table satisfying these criteria are limited, and then you show that a simple renaming of elements will reveal the group is isomorphic to $\mathbb{Z}_4$ or else the Klein-4 group.

  • 1
    $\begingroup$ Yes, this has been very helpful, thank you. Your answer is so neatly written too! The only question I have is that I believe I should write down all the group tables... Is this just 2 group tables (of what binary operation???) OR do I list more but eliminate the rest. With Cayley's table, to show commutativity, is it sufficient to use the symmetry of the diagonal line? Thanks again. PS, I have encountered Lagrange - just struggling with Algebra! $\endgroup$ – user61854 Feb 11 '13 at 21:00
  • 1
    $\begingroup$ If you have encountered Lagrange, than it suffices to argue that the only groups of order 4 have are the cyclic group $\mathbb{Z}$ (generated by one element), and a group whose non-identity elements have order 2 (since two and four divide 4). Completing any table that satisfies the properties of a group (one and only one element in each column and each row will limit the number of Cayley tables to construct. You can then show that of those, each is isomorphic to $Z_4$ or the Klein 4-group. And yes, you can show that those two non-isomorphic groups are indeed commutative by the Cayley table. $\endgroup$ – amWhy Feb 12 '13 at 1:47

There aren't that many. One of the elements will be the identity, let's call it $1$. (I am writing the operation as multiplication.) So the table looks like $$ \begin{matrix} 1 & a & b & c\\ a\\ b\\ c\\ \end{matrix} $$ (I'm not writing the row and column labels, as they are the same as the first column and row). Now what could $ab$ be? Not $a$, not $b$... And $ba$? Proceed with $ac$, $ca$, $bc$, $cb$, and then you are left with the squares. They might all be $1$, or...


This simplifies it:

If a group has 4 elements, we separate two cases:

Case1: all the elements have order 2, then the group is abelian, and $a^2=1\Rightarrow a=a^{-1}$ for all $a\in G$: $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$ the table should be easy to make in this case: there is only one.

Case2: there is an element $a\not =1$ that has order different from 2, then by LAgrange's theorem, the order must be 4, and so $G=\langle a\rangle$, and so the group is cyclic, and therefore it's abelian. We have proved that a 4-element group is always abelian. The table shouldn't be difficult in this case either.

  • $\begingroup$ You're assuming Lagrange, and chances are, at this stage, the OP has not yet encountered Lagrange. See my comment below Michael Hardy's answer. $\endgroup$ – amWhy Feb 11 '13 at 19:17
  • $\begingroup$ @amWhy I would have never imagined, Lagrange always comes really soon. I guess that then he will have to do every possible table to get to the same conclusion. $\endgroup$ – MyUserIsThis Feb 11 '13 at 21:25
  • $\begingroup$ so b^2 = a, c^2 = a in one table and in the other, b^2 = 1, c^2 = 1. $\endgroup$ – user61854 Feb 12 '13 at 16:45
  • $\begingroup$ @user61854 Yes, in my case 1, you have that: $a^2=b^2=c^2=1$, so: $ab=c, bc=a$, it's abelian, so the whole table is there. In my case 2, $a$ is the generator, so $a=a,a^2=b,a^3=c$, and you can compose to get the rest of the combinations. $\endgroup$ – MyUserIsThis Feb 12 '13 at 17:00

The order of an element must divide $4$, so it must be either $1$, $2$, or $4$. Only the identity element can have order $1$. If one element has order $4$, then it generates the whole group and you have a cyclic (hence abelian) group.

Otherwise the three non-identity elements each have order $2$.

$e=\text{the identity}$.

$a,b,c=\text{the other three}$.

$a^2=b^2=c^2=e$, and each of these is its own inverse.

So what is $ab$? It can't be $e$ since if $ab=e$ then $(ab)b^{-1} = b^{-1}=b$, and hence $a=b$. It can't be $a$ or $b$ since if $ab=b$ then $(ab)b^{-1}=bb^{-1}=e$ and so $a=e$, and similar reasoning shows it can't be $a$ (but you have to multiply on the left be the inverse).

So $ab=c$.

Similar reasoning shows $ab=ba=c$, and $ac=ca=b$ and $bc=cb=a$.

So the group is abelian.

  • $\begingroup$ Chances are, at this stage of the game, the OP hasn't encountered Lagrange. I know Fraleigh includes this exercise long before introducing Lagrange... $\endgroup$ – amWhy Feb 11 '13 at 19:15
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    $\begingroup$ @amWhy Because this is such a toy example, you don't need Lagrange; it's trivial to see that only the identity can have order 1, or that if one element has order 4 then the group must be cyclic. The only other possibility aside from the all-order-2 case is that some element (and therefore two elements) has order 3, and it's easy to eliminate this possibility. (Suppose that $a$ and $b$ have order 3, with $a^2=b$. $ca$ can't be $a$ and can't be $c$ by cancellation; it can't be $b$ because $ca=aa\implies c=a$, and it can't be the identity because then $ca=1=ba$ and again by cancellation $c=b$.) $\endgroup$ – Steven Stadnicki Feb 12 '13 at 23:33

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