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The question says that group of order $4$ cannot have an element of order $3$. We are given the hint “if it did, then, calling the elements $e, a, a^2, b$, with $a^3$=e, deduce a contradiction using the cancellation law”.

I’m not too sure on how to start the question and I would appreciate any hints to help me answer this question.

Thank you.

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    $\begingroup$ Further hint: you can’t assign a value to ab without violating the cancellation law somehow $\endgroup$ – Robo300 Dec 2 '19 at 22:23
  • $\begingroup$ Hint. Try each of the four possible things $ab$ might be. $\endgroup$ – Ethan Bolker Dec 2 '19 at 22:25
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Hint:

Consider the element $ab$. It should be equal to one of $e,a,a^2,b$. Find a contradiction for each case.

For instance: if $ab=e$, $\;b=a^{-1}=a^2$, hence the group has $3$ elements, not $4$.

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  • $\begingroup$ Why does b=$a^{-1}$ also equal $a^2$ $\endgroup$ – user728655 Dec 2 '19 at 23:08
  • $\begingroup$ Because $a$ has order $3$, i.e. $a^3=a\cdot a^2=e$. $\endgroup$ – Bernard Dec 2 '19 at 23:10
  • $\begingroup$ Thank you for making it clear. $\endgroup$ – user728655 Dec 2 '19 at 23:14
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There are some theorems which will make this immediate but I'm assuming you're not familiar with them. So consider this: the order of a group is the number of elements in the group. So, follow the hint and WLOG define $G=\{e,a,a^2,b\}$ with the order of $a$ as $3$.

Now, it follows that $ab$ must be equal to one of these elements. If $ab=e$, it implies that $a^2=b$. If $ab=a$, $b=e$, if $ab=a^2$, $a=b$ and $ab=b\Rightarrow a=e$, all of which are contradictions to the order of $G$ being $4$.

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This a direct application of Lagrange’s theorem: if $a$ were of order $3$, then $3$ would have to be a divisor of $4$.

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