# Let $G$ be a group and if $a,b \in G$ such that $a^4 =e$ and $a^2 b=ba$ then prove that $a=e$

Let $$G$$ be a group and if $$a,b \in G$$ such that $$a^4 =e$$ and $$a^2 b=ba$$ then prove that $$a=e$$.

My attempt: I know from the generators and relation of the dihedral group of order $$2n$$ is $$\langle r^n=s^2=1,rs=sr^{-1} \rangle$$ so in the given problem if I take $$r=a^2$$ and $$s=b$$ then the problem is clear that $$a^4=a$$ but in the given problem it is not given that $$b^2=e$$ so how can I prove that $$a=e$$ in the given problem.

• You can't assume it is a dihedral group. You have to use only the two relations in your title. Jun 1, 2021 at 3:14

If you know enough about conjugacy, then this can be done without writing down lots of equations. The second relation, $$a^2b=ba$$ says that $$a$$ is conjugate to $$a^2$$. So $$a^2$$ is conjugate to $$a^4$$. But $$a^4=e$$ and $$e$$ is conjugate only to itself.

If you don't know that much about conjugacy, then there's the following proof, which I got by following the preceding paragraph. Using three times that $$ba=aab$$ and once that $$aaaa=e$$, we get $$bba=baab=aabab=aaaabb=bb.$$ Cancelling $$bb$$, we get $$a=e$$.

$$a^2b=ba ... (1)$$

$$b=a^4b=a^2 (a^2 b) =(a^2 b) a= b a a= b a^2$$.

Then $$a^2=e$$.

Put it in (1), $$b=ba$$. Hence $$a=e$$.

• In your 2nd line a²(a²b)=(a²b)a...how?
– Sonu
Jun 1, 2021 at 3:25
• Use given information a^2b=ba. Replace the term in bracket with ba. Now since group multiplication is associative you can rearrange the brackets.
– Root
Jun 1, 2021 at 3:33

Start with $$a^2b = ba$$. Multiplying on the left by $$a^2$$, we get $$a^4b = a^2ba$$. However, because $$a^4 = e$$, this equation reduces to $$b = a^2ba$$. But because $$a^2ba = ba$$, the RHS of $$b = a^2ba$$ becomes $$ba^2$$. Hence, $$b = ba^2$$

Multiplying both sides on the left by $$b^{-1}$$, we have $$a^2 = e$$. But then $$ba = a^2b = eb = b$$. Multiplying on the left by $$b^{-1}$$ again, we obtain $$a = e$$.