Prove that a group of order $351$ has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.

Question

So we know $351=3^313$.

And so by the 3rd Sylow Theorem,

$n_3|13$ and $n_3 \equiv 1 \pmod{3}$

$n_{13}|27$ and $n_{13} \equiv \pmod{13}$

So $n_3 = 1, 13$ and $n_{13} = 1, 27$

So if $n_3 = 1$ we are done. It is unique and hence normal.

Similarly if $n_{13}=1$ we are done.

So now suppose $n_{13} \neq 1$

How do we show from here that a normal exists?

score 4 · Accepted Answer · answered Jun 12 '19 at 03:49

4

If $n_{13}=27$, then there are $12\times 27$ elements of order $13$. This leaves only $27$ elements of order not $13$, just enough to make one Sylow $3$-subgroup.

answered Jun 12 '19 at 03:49

Angina Seng

158,341

Why are we multiplying 12 by 27? – s_healy Jun 12 '19 at 03:53
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To exclude the identity element. – cqfd Jun 12 '19 at 03:56
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@s_healy Each Sylow $13$-subgroup has $12$ elements of order $13$. – Angina Seng Jun 12 '19 at 03:57
is it $12$ times $27$ because they all share the identity? and by elements you mean Sylow $13$ subgroups? – homosapien Oct 01 '22 at 19:57
@cqfd is my above comment correct? – homosapien Oct 01 '22 at 19:57
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what if $n_3=13$? Wheres the contradiction @cqfd – homosapien Oct 01 '22 at 20:13
@MyMathYourMath, almost: it's $12$ times $27$ because they all share the identity, *only*. – citadel Oct 02 '22 at 12:40

Prove that a group of order $351$ has a normal Sylow $p$-subgroup for some prime $p$ dividing its order.

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