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

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

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

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

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

So if $\displaystyle G$ is commutative, what about $\displaystyle gh$ 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> $\displaystyle \times$ <7> $\displaystyle \cong Z_5 \oplus Z_7$. Thus G is abelian. Since $\displaystyle Z_5 \oplus Z_7 \cong Z_{35}$ (see finitely generated abelian group), G is cyclic.

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# no of group of order35

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