# 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. Jan 3 '14 at 3:56
• @MarianoSuárez-Alvarez : I would take it as a compliment :)
– user87543
Jan 3 '14 at 4:02
• I think Sylow theorem is important in this problem. Jan 3 '14 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! Jan 3 '14 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. Jan 3 '14 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
Jan 3 '14 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
Jan 3 '14 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
Jan 3 '14 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
Jan 3 '14 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
Jan 3 '14 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 $$|C(P_7)|$$ can not be $$3$$ ,so $$=1$$ .hence $$|G/C(P_7)|=1$$ and $$P_7$$ is contained in $$Z(G)$$ ,hence $$G$$ is abelian.

• Welcome to MSE. Please use Mathjax. May 18 '17 at 6:13