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I know there are questions on this already but I wanted to check if my attempt at proving it was valid, and I couldn't find anyone else presenting the same one in any other post. My attempt is as follows:


Since there must be a unique identity, and a unique inverse for each element, I believe that without loss of generality I can let the group be $G = \{e, a, a^{-1}, b\}$ where $e$ is the identity and $a,b$ are some other elements not equal to $e$.

Then by a previous result we know that the identity element is commutative and that an element and its inverse commute with each other. This tells us that $e.x=x.e$ for all $x$ in G, and that $a.a^{-1}=a^{-1}.a$. This leaves only $a.b$ and $a^{-1}.b$ to check.

One of $a$ or $a^{-1}$ must be an inverse for $b$ and since $a,a^{-1}$ are arbitrary, say it is $a$. Then we have $a.b=b.a$. If we multiply both sides of this equality by $a^{-1}$ on the left and right we are left with $b.a^{-1}=a^{-1}.b$ which is the final possibility. We have shown that all elements in $G$ commute with each other so $G$ is abelian.


I know this proof relies on some other results which of course must be proven too but I know how to proof these, so given these are previously proven is this proof valid?

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    $\begingroup$ "One of $a$ or $a^{-1}$ must be an inverse for $b$" No. $a$ and $a^{-1}$ are each other's inverses and can't be inverses of any other elements, so $b$ must be its own inverse. You've also missed the possibility that we have $\{e, a, b, c\}$ where each element is its own inverse. $\endgroup$
    – Arthur
    Feb 13, 2018 at 11:41
  • $\begingroup$ I realise I made the mistake of assuming that if an element is its own inverse then it is the identity element. That's why I didn't realise those options were possible. Thanks. $\endgroup$
    – Eric N
    Feb 13, 2018 at 11:47
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    $\begingroup$ Also, to the community in general: I don't understand the downvotes here. There are wrong statements in the post, sure, but that's why we are here. It is clearly a good question post the way I understand the conventions here. There is a clear problem statement, followed by the OP's own attempt, quite detailed and fully readable. $\endgroup$
    – Arthur
    Feb 13, 2018 at 11:55

3 Answers 3

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All elements in such a group have order $1,2$ or $4$.

If there's an element with order $4$, we have a cyclic group – which is abelian. Otherwise, all elements $\ne e$ have order $2$, hence there are distinct elements $a,b,c$ such that $\{e,a,b,c\}= G$.

Note that, by the cancellation law, $ab\ne a$ or $b$, and similarly $ab\ne e$. Henceforth, $c=ab$. But it's also $ba$ for the same reasons. So $ab=ba$, and the group is abelian.

Note: actually this group is isomorphic to Klein's Vierergruppe $\;\mathbf Z/2\mathbf Z\times\mathbf Z/2\mathbf Z$.

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  • $\begingroup$ Sir, $ab≠e$? Is because if $ab=e$ then $ab=a^2$ so by cancellation we get $b=a$ contradiction. Am i correct? $\endgroup$ Jan 18, 2020 at 3:04
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    $\begingroup$ @AkashPatalwanshi: Yes, quite correct. You also can say that if $ab=e$, then $b=a^{-1}=a$ since $a$ has order $2$. $\endgroup$
    – Bernard
    Jan 18, 2020 at 10:04
  • $\begingroup$ It is also isomorphic to the only 2-Sylow normal subgroup $V4$ of $A4$. $\endgroup$
    – Aelx
    Nov 5, 2023 at 9:46
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Let $G=\lbrace e,a,b,c\rbrace$ and $e$ is the identity element

  • If $G$ has an element of order $4$, then $G$ is cyclic.

  • If $G$ has no element of order $4$, then $a,b,c$ are all of order $2$. That means $a^2=b^2=c^2=e$. Now $ab$ must be $c$, otherwise $ab=a$ or $ab=b$ or $ab=e$ would give a condradiction. In the same manner you can prove $ba=c=ab$, $ca=b=ac$ and $cb=a=bc$.

In all cases, $G$ is abelian. One can further prove $G$ is isomorphic to $C_4$ or $C_2\times C_2$.

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Without any appeal to orders of elements, Cauchy's Theorem of Lagrange's or type of group: assume that $G$ has $4$ elements and is not abelian. Then we can find two non-identity elements $a,b$ that do not commute, so $ab \neq ba$. Note that this implies $ab \notin \{e,a,b\}$. Because $G$ is closed under the group operation we must have $G=\{e,a,b,ab\}$. Now $ba$ belongs to this set. Since $a$ and $b$ do not commute $ba \notin \{e,a,b\}$. Hence we must have $ab=ba$, a contradiction. So $G$ is abelian after all.

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