An elementary problem in Group Theory: the unique noncyclic group of order 4 Following the advice given in this question, I have started to study Group Theory from the very basics. My reference text is Abstract Algebra by Dummit and Foote. While going through the exercises (page 24) I found one problem which required more effort than others:
Assume $G = \{1, a, b, c\}$ is a group of order $4$ with identity $1$. Also assume that $G$ has no elements of order $4$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.
I proceeded with solution as follows. Since $G$ is of even order there must be at least one element of order $2$ (this statement itself was one of the problems in the exercises and is easy to handle). From the wording of the question it seems that none of the elements $a, b, c$ is singled out with specific properties and hence $a, b, c$ must behave in exactly the same manner. Hence each of them is of order $2$. Thus $a^{2} = b^{2} = c^{2} = 1$.
Next I analyze product $ab$. Clearly it can't be $a$ (as $b$ is non-identity), or $b$ (as $a$ is non-identity) or $1$ (because $ab = 1 = aa$ implies $b = a$). Hence $ab = c$ and similarly $ba = c$. By the similar nature of all elements $a, b, c$ it follows that $bc = cb = a$ and $ca = ac = b$. Thus the operation of the group is defined properly for all elements and clearly it is abelian.
I think my solution is OK, but I am not sure. Note that I have not used the fact that $G$ has no elements of order $4$. Please let me know :
1) if my solution is correct. If not then point out the flaws.
2) if there is any better / shorter solution. If so provide hints and not solution.
The question may sound too easy / simple but I request to treat me like a beginner who has never heard of Group Theory and is reading fresh from the reference text I mentioned. Thanks in advance for your inputs.
EDIT: Thinking further about this problem I wondered what would happen if an element of order $4$ was allowed in $G$. In that case I think that the group has to be isomorphic to the group $G_{1} = \{1, 2, 3, 4\}$ with modulo $5$ multiplication as the group operation. The argument is as follows. Suppose $a$ is of order $4$ and let $b$ be its inverse then $b$ is also of order $4$. Now $c$ has to be its own inverse so that $c^{2} = 1$. Again $a^{2} \neq 1$ (as $ab = 1$), $a^{2} \neq a$ (as $a \neq 1$), $a^{2} \neq b$ (as it would mean $b^{2} = 1$ and thus $b$ would be of order $2$). Hence $a^{2} = c$ and similarly $b^{2} = c$. Next we can see that $ac \neq a$, $ac \neq c$, $ac \neq 1$ (as $ab = 1$) so that $ac = b$. Similarly $ca = b$, $bc = a$, $cb = a$. Also it is clear that this group turns out to be cyclic with both $a$ as well as $b$ as generators. And it also follows that there are the only two ($G$ in original question and $G$ in new variation) groups of order $4$ upto isomorphism.  I hope I have learnt something from the answers given for the original question and this solution is correct. Please let me know if there is any problem with this reasoning.
 A: The following is not valid:

“From the wording of the question it seems that none of the elements a,b,c is singled out with specific properties and hence a,b,c must behave in exactly the same manner.”

Luckily, this is still true:

“Thus $a^{2} = b^{2} = c^{2} = 1$.”

A correct proof will most likely use Lagrange's theorem (and make no use of Cauchy, “there is at least one element of order 2”).
Lagrange's theorem: The order of an element divides the order of the group.
Since 4 is a very small number, you could also just eliminate the possibility of order 3 directly (by hypothesis 4 is not allowed, and I hope it is clear that 5 is right out).
A: You don't need Lagrange Theorem or Cayley Theorem for this problem.
You know that $ab \neq a$ and $ab \neq b$. Then, you only have 2 options for $ab$:
$$ab=1 \, \mbox{or} \, ab=c \,.$$
We prove that $ab=1$ is not possible.
Assume by contradiction that $ab=1$, then $a,b$ are inverses to eachother, and hence the inverse of $c$ must be $c$.
Then $a^2 \neq 1, a^2 \neq a$ are obvious, while $a^2=c$ would imply that $a$ has order $4$, which is not possible. 
Thus $a^2=b$, and hence  $a^3=ab=1$. 
But then, there is no choice for $ac$. Indeed $ac \neq 1 (=ab), ac \neq b(=a^2), ac \neq c, ac \neq a$. Thus we reached a contradiction.
We proved that $ab=c$. Same way you get $ba=c$, $ac=ca=b, bc=cb=a$, and from here you can easily deduce that $a^2=b^2=c^2=1$. 
A: Here is a full solution:
This group has only one element of order 1

 ($a^1 = a$ so if $a^1=1$, then $a=1$).

This group has no elements of order 4 or greater

 (4 is outlawed by hypothesis; order 5 or greater implies that $a^1,a^2,a^3,a^4,a^5$ are 5 distinct elements of a 4 element set).

This group has no elements of order 3

 (if $a$ has order 3, then $a^2 \neq 1$ and $a^2 \neq a$, so $a^2 =b$ (or $c$, but WLOG we choose $b$). Hence $ab=ba=1$ and $b^2=a$. What about $ca$? $ca=1$ implies $c=b$. $ca=a$ implies $c=1$. $ca=b$ implies $c=a$. $ca=c$ implies $a=1$. Oh no!)

Therefore all elements have order 2, and we finish exactly as you did.
