# Math Help - Group: order 35

1. ## Group: order 35

Having troubles with this, i know that 35 = 7x5, which are both prime and subsequently 5 doesn't divide 6, so G is cyclic.

But I don't understand how to show there are elements of order 5 and 7 exactly, and that it's commutative, i.e abelian. Thanks in advance.

2. Hi

I can't see how you deduce that $G$ is cyclic from what you've written before.

Lagrange's theorem (or one of his corollary): in a finite group, the order of an element divides the order of the group.
(For what follows, I assume you don't use Sylow's theorems.)

So the elements of $G$ have order 1, 5, 7 or 35. The only of order 1 is the identity element $e$. Can the others all be of order 5? We want to show that no.

Let $x$ be an element of order $G$. The subgroup generated by $x,\ ,$ has five elements, and four different from $e$. Using Lagrange's theorem, what are their order?
Now, do you agree that in a finite group of order $n$, an element of order $n$ is a generator of the group?
So let $x,y$ be two elements of order 5 in $G$, what if $\cap \neq \{e\}$ (i.e. they have a common element)?

Therefore, if all elements in $G-\{e\}$ have order 5, all cyclic subgroups $$ have their intersections equal to $\{e\}$, and the number of elements in $G-\{e\}$ is a multiple of 4, let's say $4k$ where $k$ is a positive integer. Is $4k+1=35$ possible? No? Then contradiction, and all elements can't have order 5.
Same thing with 7.
Finally, if some element $a\in G$ has order 35 (i.e. is a generator of $G$), what about $a^5$ and $a^7$ ?

I guess that we've proved that $G$ contains at least an element $g$ of order 5 and an element $h$ of order 7.

Now, what we want to show is that given a $G$ of order 35, if $G$ is abelian, then it is cyclic.

So if $G$ is commutative, what about $gh$ order ? Conlude

3. This is a proof that uses a Sylow theorem.

Lemma 1. If there is only one Sylow p-subgroup P in G, then P is normal in G.

By Sylow's theorem, G has subgroups of order 5 and order 7.

If the order of G is pq, then the number of Sylow q-subgroup is kq+1 and divides pq.
We see that there exists only one Sylow 5-subgroup and one Sylow 7-subgroup in G. Thus, both are normal in G by lemma 1.

Since both <5> and <7> are normal in G and intersects only at e in G, G = <5> $\times$ <7> $\cong Z_5 \oplus Z_7$. Thus G is abelian. Since $Z_5 \oplus Z_7 \cong Z_{35}$ (see finitely generated abelian group), G is cyclic.