Group of order $1575$ problem

Problem

Let $G$ be a group with $|G|=1575$. If $H \lhd G$ and $|H|=9$, then $H \subseteq Z(G)$.

What I've done so far is $|G|=1575=3^25^27$. I consider $G$ acting on $H$ by conjugation, or, in other words, I consider the morphism $$\phi: G \to Aut(H)$$$$g \to ghg^{-1}, \forall h$$

Since $H$ is normal, it is clear that $\phi(g) \in Aut(H)$. Now, $$H=\coprod_{h\in H} \mathcal O_h,$$ where $\mathcal O_h=\{x \in H: ghg^{-1}=x\}$. If I could show that each of these orbits has one element, then it easy to see that $h \in Z(G)$ for all $h \in H$.

One can define a bijection between each $\mathcal O_h$ and $G/ G_h$, where $G_h=\{g \in G :ghg^{-1}=h\}$, so $1=|\mathcal O_h|=\dfrac{|G|}{|G_h|}$.

Then, $|G_h|=|G|$, at this point I got completely stuck, another thing I know is that if $|H|=3^2$ and $H$ is a normal subgroup, then $H$ is the only $3$-Sylow subgroup of $G$.

I would appreciate some suggestions, thanks in advance.

• Do you know the normalizer/centralizer theorem (or N/C theorem)? Sep 27 '14 at 18:12
• Hmm, not really, what does it say? (I can look it up in some textbook). Sep 27 '14 at 18:16
• Do you mean to use the fact that $N(H)/C(H) \cong Aut(H)$? If the normalizer and centralizer are the same, then the cardinal of $Aut(H)$ is the identity, and this is equivalent to $H \subset Z(G)$. Sep 27 '14 at 18:21
• Hint: $|H|=9$ implies that either $H\cong \Bbb{Z}_9$ or $H\cong\Bbb{Z}_3\times\Bbb{Z}_3$. Their automorphism groups are $\Bbb{Z}_9^*$ and $GL_2(\Bbb{Z}_3)$ respectively. Do you know the orders of those groups? Sep 27 '14 at 18:21

Read Tobias comment: the N/C (*) theorem tells us that

$$N_G(H)/C_G(H)=G/C_G(H)\cong T\le\;\text{Aut}\,(H)$$

But since $$\;H\cong C_3\times C_3\;$$ or $$\;H\cong C_9\;$$ , we have that $$\;|\text{Aut}\,(H)|\in\{6\;,\;48\}\;$$ , and since

$$\left|G/C_G(H)\right|\ \left|\ \frac{1575}9\right.=175=5^2\cdot7$$

so it must be that $$\;G/C_G(H)=1\iff C_G(H)=G\iff H\le Z(G)\;$$

(*) The N/C theorem tells us that we have a homomorphism

$$\phi:N_G(H)\to\;\text{Aut}\,(H)\;,\;\;\phi(x):=\varphi_x$$

with $$\;\varphi_x(h):=h^x:=x^{-1}hx\;,\;\;\forall\,h\in H\;$$

and then the first isomorphism theorem gives us

$$N_G(H)/\ker\phi\cong T\le\;\text{Aut}\,(H)$$

for some subgroup $$\;T\;$$ of the automorphism group of $$\;H\;$$

• I have revised a typo in the fourth line.
– Bach
Sep 25 '19 at 14:43

Case 1: $H$ is cyclic, generated by $a$. We will show $Z_G(a) = G$, which would imply $H \subset Z(G)$. Since $H$ is normal, $gag^{-1} = a^k$ for some $1 \leq k \leq 9$. Clearly, $|H| = 3^2$ divides $|Z_G(a)|$. By applying the conjugation $i$ times, we see that $g^{i}ag^{-i} = a^{k^i}$ (where we will choose $g$ suitably). Let $P_5$ and $P_7$ be Sylow subgroups. If $g \in P_5$, we put $i = 25$ and obtain $g^{25}ag^{-25} = a = a^{k^{25}}$, which tells us $k^{25} \equiv 1 \mod 9$. Some casework yields $k \equiv 1 \mod 9$. Similarly, putting $i = 7$ for $g \in P_7$ yields $k \equiv 1 \mod 9$.

This tells us $P_5, P_7, H$ are all subgroups of $Z_G(a)$ and thus their orders must divide $|Z_G(a)|$, which must be 1575.

Case 2: $H$ is generated by $a, b$, with $a^3 = b^3 = 1$. If $gag^{-1} = a^k$, then $gbg^{-1} = b^j$, and this is similar to case $1$ and we get $k^{25} \equiv 1 \mod 3$, $k^7 \equiv 1 \mod 3$, both of which imply $k \equiv 1 \mod 3$.

If $gag^{-1} = b^j$, and $gbg^{-1} = a^k$ (here $j, k \in \{1, 2 \}$), then applying the conjugation $4$ times yields $g^4 a g^{-4} = a$ (verify this is true mod 3). Since $4$ is coprime to $5$ and $7$, the fourth power map is surjective in $P_{25}$ and $P_7$, and we can finish up this case in a manner similar to case 1.