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How can we prove that "Every Group of order $216$ contains a normal subgroup of order $27$ or $9$ ?".

MY TRY: $216=2^3.3^3$.So possible order of sylow $3$ subgroup is $1$ or $4$ and similarly for sylow $2$ subgroup is of order $1,3,9\,\text{or}\,27$.Then how to proceed $?$Thank you

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  • $\begingroup$ If there is $1$ Sylow $3$-subgroup then you are done, so assume there are $4$. So there is a homomorphism from $G$ to $S_4$, image must be $A_4$ or $S_4$, kernel has order $18$ or $9$. $\endgroup$ – Derek Holt Jan 24 '17 at 6:57
  • $\begingroup$ I think in 2nd line you mean$A_4 $or $K_4$ $\endgroup$ – MatheMagic Jan 24 '17 at 7:00
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    $\begingroup$ I meant what I wrote: $A_4$ or $S_4$. The image has to have order divisible by $3$, because one Sylow $3$-subgroup cannot normalize another one. $\endgroup$ – Derek Holt Jan 24 '17 at 8:49
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Suppose there are $\;4\;$ Sylow $\;3\,-$ subgroups of order $\;3^3=27\;$ , which means there is a subgroup of index $\;4\;$ in $\;G\;$ , namely: $\;N_G(P_3)\;,\;\;P_3=\;$ any Sylow $\;3\,-$ subgroup.

But making $\;G\;$ act on the four cosets of this subgroup, we get a homomorphism $\;\phi: G\to S_4\;$ characterized by

$$\ker\phi=\bigcap_{g\in G}P_3^g =\text{ the core of}\;\;P_3=$$

the maximal normal subgroup of $\;G\;$ which is contained in $\;P_3\;$ . Clearly then $\;\phi\;$ cannot be injective (why? Look at orders...), and in fact its order must be such that, by the first isomorphism theorem

$$G/\ker\phi\cong K\le S_4$$

Well, this already shows there's a non-trivial normal subgroup in $\;G\;$ , and now calculate what the possible order of it can be by the above.

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