I googled my question, nothing appeared. My book says that group of order 63 is Abelian. The way I see it is perfectly possible that it has 7 Sylow 3 subgroups and one Sylow 7 subgroup. Please help!
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3it is a group of order $p^2q$ – Asinomás Jan 07 '16 at 16:27
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Note that "group of order $63$" may imply Abelian without necessarily requiring all such groups to be isomorphic. – hardmath Jan 07 '16 at 16:32
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3$\mathbb{Z}_7\rtimes \mathbb{Z}_9$ and $\mathbb{Z}_7\rtimes (\mathbb{Z}_3\times\mathbb{Z}_3)$ are non-abelian groups of order $63$. – Tristan Phillips May 19 '18 at 17:05
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If that is the case, what are the (maximal) numbers of elements of orders
- is either $3$ or $9$?
- is equal to $7$?
Quang Hoang
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I'll try to write an answer describing groups of order $63=3^2\cdot7$.
(If there is any mistake please let me know)
It is $n_7|9,\ n_7\equiv 1\pmod7\Rightarrow n_7=1$ so there is a unique Sylow $7-$subgroup $P_7\lhd G$.
It is also $n_3|7,\ n_3\equiv 1\pmod3\Rightarrow n_3=1,7$. Let $P_3$ be a Sylow $3-$subgroup. It is $P_3=9$ so $P_3\cong \mathbb{Z}_9$ or $\mathbb{Z}_3\times\mathbb{Z}_3$
- If $n_3=1$ then $P_3\lhd G$ and $|P_3P_7|=|G|$ with $P_3P_7\leq G$ hence $G=P_3\times P_7\cong \mathbb{Z}_9\times \mathbb{Z}_7$ or $\mathbb{Z}_3\times \mathbb{Z}_3\times \mathbb{Z}_7$.
- If $n_3=7$ it is $G=P_7\rtimes_{\phi}P_3$ where $\phi:P_3\to Aut(P_7)$. Since $P_3\not\lhd G$ it follows that $\phi$ is non-trivial. Hence (since $Aut(\mathbb{Z}_7)\cong \mathbb{Z}_6$) $\phi$ maps $P_3$ to the unique subgroup of $\mathbb{Z}_6$ of order $3$. So $G\cong \mathbb{Z}_7\rtimes_{\phi}\mathbb{Z}_9$ or $G\cong \mathbb{Z}_7\rtimes_{\phi}[\mathbb{Z}_3\times \mathbb{Z}_3]$.
In the first case $\phi:P_3=\langle x\rangle \to Aut(P_7)=\mathbb{Z}_6=\langle \tau\rangle,\phi(x)=\tau,\ \tau(a)=a^2$ and in the second case $\phi:\langle x\rangle\times\langle y\rangle\to Aut(P_7)=\mathbb{Z}_6=\langle \tau\rangle,\ \phi(x)=\tau,\phi(y)=1,\tau(a)=a^2$
1123581321
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1You might want to say more about the last case in your answer: the reason you only get one isomorphism class is because your Sylow 3-group is completely reducible: the kernel of $\phi$ has a complement. – Steve D Aug 29 '20 at 07:58
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@Steve D Thanks for the comment. I thought it was obvious by the symmetry of $x,y$ but I'll try to see what you mean – 1123581321 Aug 29 '20 at 08:03
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Well I can define a new $\tilde{\phi}$ with $\tilde{\phi}(x)=\tilde{\phi}(y)=\tau$. But the semidirect product I get is isomorphic to yours. – Steve D Aug 29 '20 at 08:06
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@Steve D So if I understand correctly you say that $P_3=ker\phi\times Q$ and this gives me the fact that the two groups are isomorphic, but I can't see why. Let me think for a bit – 1123581321 Aug 29 '20 at 08:10
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Since $f:\langle x\rangle \times \langle y\rangle \to \langle x\rangle \times \langle y\rangle$ with $f(x)=y,f(y)=x$ is an automoprhism of $P_3$ I think that this is enough to explain why the semidirect products $P_7 \rtimes_{\phi} P_3,\ P_7 \rtimes_{\phi\circ f} P_3$ are isomorphic. I try to see whether this works for your $\bar{\phi}$ – 1123581321 Aug 29 '20 at 08:18
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@Steve D I think that $f: \langle x\rangle \times \langle y\rangle \to \langle x\langle \times \langle y\rangle$ with $f(x)=x,f(y)=yx$ is an isomorphism and $\phi\circ f=$ yours $\bar{\phi}$. What do you think? – 1123581321 Aug 29 '20 at 08:24
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