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I am given this problem, and the solution reasons differently from me and obscurely, to me

Let $G$ have order $30$. Show that it is not simple.

The solution is apparently

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So, what I don't undestand is, say $n_3=10$ for a contradiction. This tells me that there are $10$ distinct Sylow $3$-subgroups. Now Since each have prime order $3$, they are all cyclic. By distinctness, each need to have a different generator or they will have identical elements. This simply tells me that $G$ has at least $10$ distinct elements that have order $3$. Period.

Why $20$? In a similar manner, I don't understand where the $4$ came in for $n_5$. It says "we use Lagrange" but how? Take $P_j$ for the $j$th Sylow $5$ subgroup then simply, Lagrange tells us merely that $30=6 \cdot |P_j|$. So there are $6$ different left or right cosets of $P_j$ by $G$. Period.

I can infer nothing more or anything close to what the solution is saying, very obscure. Please help

Have been linked to an almost identical problem, and I read through the solutions given. Now, I think I would like to still ask "why does the nontrivial elements of a cyclic group always generate the group?" Trying it out with small groups it turns out true but is there a rigorous proof of it?

And also, still, why is Lagrange showing up here?

Melba1993
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    Not of any cyclic group: just for the ones of prime order (such as $5$). In general, if $C_n$ is a cyclic group and $g$ is one its generators, the generators of $C_n$ are $g^{k}$ for $k$ coprime with $n$. Of course, this yields that two cyclic subgroups of order $5$ must intersect trivially. –  May 14 '16 at 13:24

1 Answers1

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1) Note that if $\{1,a,a^2\}$ is a group then both $a$ and $a^2$ are generators.

2) In a group has prime order $p$ then every element must have order dividing $p$ by Lagrange. So every element has order $p$ except the identity. So every element apart from the identity in a group of prime order is a generator.

Edit in responce to comment:

Consider an element g of a group G such that g does not divide the order of G. Then calculate the order of the group generated by g. Notice the order of this group does not divide the order of G.

icecrystal
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  • I'll just remember that a cyclic group of prime order has every element except the identity as sa generator. But for the second point, my only understanding of Lagrange is, $|G|=|cosets| \cdot |H|$. Now $p=|H| \cdot |cosests|$, how does this imply "every element must have element of order dividing $p$?? It tells me the subgroups are trivial, but isn't that it? – Melba1993 May 14 '16 at 13:46