Prove that a group of order 36 must have a normal subgroup of order 3 or 9.

Let n2 be the number of 2-Sylow subgroups of G (with |G|=36). Then n must be 1 or 3. Let n3 be the number of 3-Sylow subgroups of G. then n3=1 or n3=4 if n3=1 we have 1 3-sylow group of order 9. and it is also a normal group(from sylow theorem ) if n2=1 there is normal group of order 4 but I cant show normal group of order 3.

  • $\begingroup$ If not, the 3-Sylows would be core-free, so your group would embed in $S_4$. $\endgroup$
    – Martino
    Oct 30 '12 at 17:45
  • $\begingroup$ See this link math.stackexchange.com/q/57754/8581. $\endgroup$
    – Mikasa
    Oct 30 '12 at 17:49

Assume that $G$ has $4$ Sylow $3$-groups, as you noted $n_3=(1+3k)|4, n_3\neq1$. Defining the conjugation action on the set of these $4$ Sylow $3$-groups, we have the induced homomorphism $\phi: G\longrightarrow S_4$. So $\frac{G}{\ker\phi}\hookrightarrow S_4$, but $3^2$ divides |G|=36 and does not divide $|S_4| = 24$, so $\ker\phi\neq 1$ and of course $\ker\phi\neq G$. This means that there must be non-trivial kernel, which is a non-trivial normal subgroup of G.

  • $\begingroup$ Is the kernel guaranteed to have order 3? I was trying to solve this same problem, but I couldn't get past this part: the image of $\phi$ has to divide both 24 and 36, so it can be any factor of 12, but for the kernel to be 3 we need the image to be exactly 12. Why isn't it, say, 6? $\endgroup$
    – coldnumber
    Aug 3 '16 at 15:17

The normal subgroup of order 3 in a group of order 36 is guaranteed. By Sylow theorem the subgroup of order 9 is possible. Also index of the subgroup is 4. And 36 does not divide 4!=24. Hence there is a non trivial normal subgroup contained in a subgroup of order 9.By lagrange theorem those normal subgroup order is 3. Hence the normal subgroup of order 3 exist in group having order 36.


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