# "Every Group of order $216$ contains a normal subgroup of order $27$ or $9$

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

• 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 $9$. – Derek Holt Jan 24 '17 at 6:57
• I think in 2nd line you mean$A_4$or $K_4$ – MatheMagic Jan 24 '17 at 7:00
• 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. – Derek Holt Jan 24 '17 at 8:49

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.