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I'm trying to prove that if $|G|=30=2\cdot 3\cdot 5$ then $G$ has a normal $3$-Sylow or a normal $5$-Sylow.

By the Sylow theorems, we would argue by contradiction in order to prove it cannot be $n_3=10$ and $n_5=6$ ($n_p$ is the number of $p$-Sylows).

Here: Prove that a group of order 30 has at least three different normal subgroups

In the second answer it is stated that if $n_3=10$ and $n_5=6$, then counting the number of distinct elements of these subgroups we have $10\cdot(3-1)+6\cdot(5-1)+1>30$.

However, I don't understand why this number is $10\cdot(3-1)+6\cdot(5-1)+1$. For example, is he saying that the number of elements of the $3$-Sylows (without the identity) is $10\cdot (3-1)$? Can anyone explain why? Wouldn't it be true only if they're disjoint?

Thank you.

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Yes, any two of the mentioned subgroups (10 subgroups of order $3$ and 6 of order $5$) are necessarily disjoint - meet only in the unit -, because their intersection must be a proper subgroup of both, so it can only be the trivial group as $3$ and $5$ are primes.

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  • $\begingroup$ Oh, now I understand. Thank you very much! $\endgroup$ – Talexius Jun 14 '16 at 21:00

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