# Group of order $1575$ having a normal Sylow $3$ subgroup is abelian.

Question is to prove that :

If a group $$G$$ with $$|G|=1575=3^2\cdot5^2\cdot 7$$ has a normal Sylow $$3$$ subgroup then :

• Sylow $$5$$ subgroup is normal
• Sylow $$7$$ subgroup is normal

In this situation, Prove that $$G$$ is abelian.

All i can do is :

If Sylow $$3$$ subgroup, Sylow $$5$$ subgroup, Sylow $$7$$ subgroup is normal then $$G$$ is abelian.

Notation : $$P_3$$ for Sylow $$3$$ subgroup ;$$P_5$$ for Sylow $$5$$ subgroup; $$P_7$$ for Sylow $$7$$ subgroup

We know that :

For $$H\leq G$$ the quotient group $$N_G(H)/C_G(H)$$ is isomorphic to a subgroup of $$\text{Aut(H)}$$.

• As $$P_3\unlhd G$$ we have $$N_G(P_3)=G$$
• As $$P_5\unlhd G$$ we have $$N_G(P_5)=G$$
• As $$P_7\unlhd G$$ we have $$N_G(P_7)=G$$

Thus, we will have

The quotient group $$G/C_G(P_i)$$ is isomorphic to a subgroup of $$\text{Aut(P_i)}$$ for $$i=3,5,7$$.

In case of $$P_3$$ we have $$G/C_G(P_3)\cong M \leq \text{Aut(P_3)}$$

Now, $$|P_3|=3^2$$ so, $$|\text{Aut(P_3)}|=3(3-1)=6$$

As $$C_G(P_3)\leq G$$ we see that $$|C_G(P_3)|$$ divides $$|G|$$ with the condition $$|G/C_G(P_3)|$$ divides $$6$$.

As $$P_3$$ is abelian we have $$H\leq C_G(P_3)$$ so, $$G/C_G(P_3) \leq G/P_3$$

i.e., $$G/C_G(P_3)\leq G/P_3$$ i.e., $$|G/C_G(P_3)|$$ divides $$|G/P_3|=5^2\cdot7$$

we already have a condition that $$|G/C_G(P_3)|$$ divides $$6$$.

But, $$6$$ and $$5^2.7$$ do not have a common factor other than $$1$$ so, $$|G/C_G(P_3)|=1$$

i.e., $$C_G(P_3)=G$$ i.e., $$P_3\leq Z(G)$$.

In case of $$P_5$$ we have $$G/C_G(P_5)\cong M \leq \text{Aut(P_5)}$$

Now, $$|P_5|=5^2$$ so, $$|\text{Aut(P_5)}|=5(5-1)=20$$

As $$C_G(P_5)\leq G$$ we see that $$|C_G(P_5)|$$ divides $$|G|$$ with the condition $$|G/C_G(P_5)|$$ divides $$20$$.

As $$P_5$$ is abelian we have $$H\leq C_G(P_5)$$ so, $$G/C_G(P_5) \leq G/P_5$$

i.e., $$G/C_G(P_5)\leq G/H$$ i.e., $$|G/C_G(P_5)|$$ divides $$|G/P_5|=3^2\cdot7$$

we already have a condition that $$|G/C_G(P_5)|$$ divides $$20$$.

But, $$20$$ and $$3^2.7$$ do not have a common factor other than $$1$$ so, $$|G/C_G(P_5)|=1$$

i.e., $$C_G(P_5)=G$$ i.e., $$P_5\leq Z(G)$$.

In case of $$P_7$$ we have $$G/C_G(P_7)\cong M \leq \text{Aut(P_7)}$$

Now, $$|P_7|=7$$ so, $$|\text{Aut(P_3)}|=(7-1)=6$$

As $$C_G(P_7)\leq G$$ we see that $$|C_G(P_7)|$$ divides $$|G|$$ with the condition $$|G/C_G(P_7)|$$ divides $$6$$.

As $$P_7$$ is abelian we have $$H\leq C_G(P_7)$$ so, $$G/C_G(P_7) \leq G/P_7$$

i.e., $$G/C_G(P_7)\leq G/P_7$$ i.e., $$|G/C_G(P_7)|$$ divides $$|G/P_7|=3^2\cdot7$$

we already have a condition that $$|G/C_G(P_7)|$$ divides $$6$$.

Now there is a hitch....

I can not use same arguments as i have used for $$P_3$$ and $$P_5$$ as $$3$$ does divide $$6$$ and $$3^2.7$$.

So, I can not immediately conclude $$|G/C_G(P_7)|=1$$ i.e., $$G=C_G(P_7)$$ i.e., $$P_7\leq Z(G)$$.

Assuming I have Proved $$P_i\leq Z(G)$$ for $$i=3,5,7$$

It would not take much time to conclude $$G=Z(G)$$

As $$P_i\cap P_j =\{e\}$$ for $$i\neq j$$ and $$i,j\in \{3,5,7\}$$ we see that $$G=\langle P_3,P_5,P_7\rangle$$

But then $$P_i\leq Z(G)$$ for $$i=3,5,7$$ i.e., $$G =\langle P_3,P_5,P_7\rangle \leq Z(G)$$

i.e., $$G=Z(G)$$ which means $$G$$ is abelian.

So, I am done on more than $$40$$ percent of the problem leaving possibility of $$|G/C_G(P_7)|=3$$ and I do not really understand how to make use of $$P_3$$ being Normal to conclude $$P_5$$ and $$P_7$$ are Normal.

I am not sure of the procedure but I have something to say on $$P_5$$ being normal :

As $$P_3$$ is normal I can consider $$G/P_3$$ with $$|G/P_3|=5^2.7$$

So, If i see this group as something named $$M$$ This subgroup have

• a Sylow $$5$$ subgroup and
• a Sylow $$7$$ subgroup.

But the condition $$n_5=1+5k$$ dividing $$7$$ gives only possibility that $$n_5=1$$

Which means that Sylow $$5$$ subgroup of $$M$$ (I wish i could make this as $$P_5$$) is normal.

With similar reason $$n_7=1+7k$$ dividing $$5^2$$ gives only possibility that $$n_7=1$$

Which means that Sylow $$7$$ subgroup of $$M$$ (I wish I could make this as $$P_7$$) is normal.

So I guess I am done up to $$60$$ percent.

I am not so sure how to make use of (believing that I can make use of) Sylow $$5$$ subgroup, Sylow $$7$$ subgroup of $$M$$ being normal to conclude $$P_5$$ and $$P_7$$ are normal.

I would be thankful if some one can help me to clear two gaps in this :

• How to get rid of $$|G/C_G(P_7)|=3$$.
• How to make use of Sylow $$5$$ subgroup, Sylow $$7$$ subgroup of $$M$$ being normal to conclude $$P_5$$ and $$P_7$$ are normal.

Thank you.

P.S : Please give "just hints". Do not write whole answer at once. This is a "request". Thank you :)

• I love your percentages. Commented Jan 3, 2014 at 3:56
• @MarianoSuárez-Alvarez : I would take it as a compliment :)
– user87543
Commented Jan 3, 2014 at 4:02
• I think Sylow theorem is important in this problem. Commented Jan 3, 2014 at 4:09
• Are you assuming that both Sylow 3- and 5-subgroups are cyclic? When you calculate the order of their automorphism groups I think you have to consider the cases of being elementary abelian too! Commented Jan 3, 2014 at 6:31
• In that case you have to consider $P_3 \cong C_3 \times C_3$ too, whence Aut$(P_3) \cong SL(2,3)$. Same for the prime 5. Commented Jan 3, 2014 at 7:08

Hint: Pull back Sylow subgroups of $G/P_3$, which are normal, and use the fact that if a Sylow subgroup is normal then it is characteristic.

• I do not get your point.. could you please extend this a bit...
– user87543
Commented Jan 3, 2014 at 4:34
• The preimage of Sylow 5(resp. 7)-subgroup of $G/P_3$ is normal in $G$ and it has a normal, hence characteristic Sylow 5(resp. 7)-subgroup. Then it must be (why?) a Sylow 5(resp. 7)-subgroup of $G$.
– user33321
Commented Jan 3, 2014 at 4:39
• I am sorry for late reply... There was problem with my computer... we consider $\eta : G\rightarrow G/P_3$ then you are asking me to cvonsider $\eta^{-1}(M_1)$ where $M_1$ is that normal sbgroup of $G/P_3$ but why would $\eta^{-1}(M_1)$ is normal if $\eta$ is just a homomorphism(surjective)
– user87543
Commented Jan 3, 2014 at 5:55
• Oh yes.... I got it... but why do we need being characteristic... :O I am sorry I am messing it up!
– user87543
Commented Jan 3, 2014 at 11:01
• OH my Bad.. I got it now!!! Perfect!! Thank you!! could you please help me to look at the case of $|C_G(P_7)|=3$
– user87543
Commented Jan 3, 2014 at 11:16

If $$|G/C(P_7)|=3$$ then $$|C(P_7)|=3*5^2*7$$, now $$3^2$$ divides $$|Z(G)|$$ and $$5^2$$ divides $$|Z(G)|$$. So $$|Z(G)|$$ is greater or equal to $$3^2*5^2$$, so $$|Z(G)|=3^2*5^2$$ or $$3^2*5^2*7$$ but $$Z(G)$$ is subgroup of $$C(P_7)$$ so $$|Z(G)|$$ divides $$|C(P_7)|$$, but in either case $$|Z(G)|$$ can not divide $$|C(P_7)|$$, so $$|C(P_7)|\ne 3$$, which implies $$|C(P_7)|=1$$.Hence $$|G/C(P_7)|=1$$ and $$P_7$$ is contained in $$Z(G)$$ ,hence $$G$$ is abelian.

• Welcome to MSE. Please use Mathjax. Commented May 18, 2017 at 6:13