Note-This is not a duplicate, I do not need answer to the whole question but I am looking for a small explanation.
First thing is that after reading the question I am having a doubt. and then I cannnot understand some part of a proof.
I was asked that given an abelian group $G$ of order $35$ and every element of $G$ satisfies the equation $x^{35}=e$.Prove that $G$ is cyclic.
Doubt in the Question What is the necessity of giving that every element of $G$ satisfies the equation $x^{35}=e$.Isn't is obvious. if $x \in G$ implies that if $|x|=m$ then $m|35$ hence it is definitely true that $x^{35}=e$ for every element $x\in G$
Coming to the Part which I couldn't understand.I have written it in bold I referred to a solution which is like this...
Proof-a nonidentity element of $G$ must have order $5, 7$ or $35$. We may assume that $G$ has no element of order 35. Since $34$ is not a multiple of $ \phi \left(5\right) = 4,$ not all of the nonidentity elements can have order $5$. Similarly, not all of them can have order $7$. So, $G$ has elements of orders both $5$ and $7$. Say, $|a| = 5$ and $|b| = 7.$ Then, since $\left(ab\right)^5 = b^5 \neq e$ and $\left(ab\right)^7 = a^7= a^2 \neq e,$ we must have $|ab| = 35,$ a contradiction.
Please give me the explanation for that statement.That is enough to help me Thanks
Yes, the condition $x^{35}=e$ is redundant by Lagrange's theorem.
The powers of any element of order $5$ form a subgroup of order $5$, which includes the identity and 4 other elements, all of which have order $5$. It is easy to see that for the 34 non-identity elements, all those of order $5$ can only be in one of these subgroups (because when an element is in a subgroup so are all its powers, essentially). So if all the non-identity elements had order $5$, they'd split 34 into equal sets of size $4$, which is impossible since $4 \not\mid 34$. Similarly, $6 \not\mid 34$, so they can't split into subgroups that all have order $7$.