# Group of order 4 cannot have element of order 3

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.

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

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 $$4$$.

• Why does b=$a^{-1}$ also equal $a^2$ – user728655 Dec 2 '19 at 23:08
• Because $a$ has order $3$, i.e. $a^3=a\cdot a^2=e$. – Bernard Dec 2 '19 at 23:10
• Thank you for making it clear. – user728655 Dec 2 '19 at 23:14

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 $$4$$.

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