# doubt in proving that an abelian group of order 35 is cyclic.

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

• There is a general rule for groups $G$ of order $q\cdot p$ where $q < p$ and both are prime numbers. If $q \nmid p-1$, then $G$ is always cyclic. You could try and prove this proposition. Commented May 25, 2018 at 21:10

Suppose all non identity elements have order $5$, and consider the sets $X_a=\bigl\{a, a^2,a^3,a^4\bigr\}$ for all $a\in G$, $a\ne e$. Once you've removed the duplicates, these sets make up a partition of $G\setminus\{e\}$, so $\;34=\bigl|\mkern1mu G\setminus\{e\}\mkern1mu\bigr|$ is a multiple of $4$. Contradiction.
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 $7$.
• Thank You. Can this solve for the general case where instead of 35...we take $p_1p_2....p_n$ where all $p_i$ are distinct prime numbers.Can we move ahead with same reasoning. In this case I am getting $p_i-1|p_1p_2....p_n-1$.I do not think it will hold when we work on 33 instead of 35. Commented May 24, 2018 at 9:35
• Yes, this won't work in general. In the case of distinct primes, you can apply Cauchy's theorem to produce an element with order $p_i$ for each $i$, and their product then has order $p_1p_2\dotsm p_m$ and so is a generator for $G$. Commented May 24, 2018 at 10:57