No simple groups of order $90$

Question

So I was trying to study the Sylow theorems.

I tried to prove that there are no simple groups of order $90$.

I proved that if $n_{5}=1$ or $n_{3}=1$ then it has a normal subgroups (meaning $G$ is not simple). Now I'm trying to figure out what happens if $n_{5}=6$ and $n_{3}=10$. I saw the following statements:

If group $G$ has 6 sylow-5 subgroups then $G$ has $6$ subgroups of order $5$, when each one of them has $4$ elements of order $5$. Every two groups are equal or have trivial Intersection Then $G$ has $6\cdot 4=24$ elements of order $5$.

I do understand why $G$ has $6$ subgroups of order $5$ but why each one of them has $4$ elements of order $5$? Also, what does it mean "Every two groups are equal or have trivial Intersection"?

(I read the previous threads).

a group of order $5$ is cyclic, containing the identity element and $4$ other elements, which are of order $5$ — J. W. Tanner, Feb 17 '19 at 16:12
The order of an element of a group has to divide the order of the group, and for prime order groups, there aren't many possibilities... — J. W. Tanner, Feb 17 '19 at 16:17

score 1 · Answer 1 · answered Feb 17 '19 at 16:27

The Sylow 5-subgroups are cyclic, because groups of prime order are cyclic. They each contain the identity element and four other elements, which are each of order 5. The intersection of two Sylow 5-subgroups is either the entire subgroup or trivial (meaning just the identity). (This is because, as subgroups, they must contain the identity. If they contain another element in common, then they contain the subgroup generated by that element, which is five elements -- the entire subgroup.)

No simple groups of order $90$

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