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I am working on an exercise where, so far, I have constructed a group $G = C_7 C_6 = \big\{n'h'\;|\;n' \in C_7,\;h' \in C_6\big\}$ defined by the relations $$\big\{n^7 = h^6 = 1,\;hnh^{-1} = n^2\big\}$$ Indeed, $G$ is the extension of $C_6$ by $C_7$ such that the generator of $C_6$ acts by conjugation on $C_7$ as an automorphism of order 3. Now, since $|G| = 2\cdot 3\cdot 7$, we know by the Sylow theorems that there is a unique Sylow 7-subgroup, 1 or 7 Sylow 2-subgroups, and 1 or 7 Sylow 3-subgroups. It's not clear to me how we can determine that there is exactly 1 Sylow 2-subgroup and 7 Sylow 3-subgroups. I have tried a proof by contradiction, but I can't get things to work out properly.

Update: I have proven that there is exactly one Sylow 2-subgroup and 7 Sylow 3-subgroups. However, my method was sort of by brute force. That being said, I'm still curious if there is a quick way to determine this result.

slowspider
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    But it isn't acting as an automorphism of order $2$, it is acting as an automorphism $\tau$ of order $3$: $\tau:n \mapsto n^2$, $\tau^2:n \mapsto n^4$, $\tau^3:n \mapsto n^8 = n$. – Derek Holt Oct 07 '21 at 07:29
  • My mistake, that is what I meant to write. – slowspider Oct 07 '21 at 13:06
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    So $h^3 \in Z(G)$, and since $h^3$ has order $2$, it generates the unique Sylow $2$-subgroup of $G$. But the Sylow $3$-subgroup $\langle h^2 \rangle$ of $G$ is not normalized by $n$, so it is not normal in $G$ and hence there must be $7$ Sylow $3$-subgroups. – Derek Holt Oct 07 '21 at 14:58
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    All right, that makes sense. However, how do we know that the Sylow 2-subgroup is unique? – slowspider Oct 07 '21 at 16:32
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    The notation $C_6C_7$ is rather unfortunate. What you have is a semidirect product of $C_7$ by $C_6$, so it should be $C_7\rtimes C_6$ or $C_6\ltimes C_7$. (Note that whether this is "an extension of $C_6$ by $C_7$" or "an extension of $C_7$ by $C_6$" depends on what definitions you use; both are correct. See here for a discussion. – Arturo Magidin Oct 07 '21 at 17:52
  • It is unique because it is in the centre of the group as I said in my previous comment. – Derek Holt Oct 07 '21 at 19:57
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    @ArturoMagidin I have had this problem a lot. Broadly speaking, if $G$ is a group then it is an extension of the kernel by the quotient. If $M$ is a module, then it is an extension of the quotient by the kernel. Group theorists use the one, representation theorists use the other. People who sit uneasily between the two (e.g., Marty Isaacs, myself) have to remember which hat they are wearing at the time. – David A. Craven Oct 07 '21 at 21:55

1 Answers1

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The group elements can be written as $n^a h^b$ with $a\in\{0,\ldots,6\}$ and $b\in\{0,\ldots,5\}$. To do multiplications in that representation, we need to know how we can interchange powers of $h$ with powers of $n$.

So let us investigate that first. By $hnh^{-1} = n^2$ we have $hn = n^2 h$. Iteratively, we get that $h n^a = n^{2a} h$ for all integers $a$. (For negative integers $a$, use that inverting the equality $hnh^{-1} = n^2$ yields $hn^{-1} = n^{-2} h$.) Similarly, iterating over $b$ gives $h^b n^a = n^{2^b a} h^b$ for all integers $a, b$.

Now let us examine the obvious $3$-Sylow subgroup $P_3 = \langle h^2\rangle = \{1, h^2, h^4\}$ and the obvious $2$-Sylow subgroup $P_2 = \langle h^3\rangle = \{1,h^3\}$.

We compute $$n h^2 n^{-1} = n (h^2 n^{-1}) = n (n^{2^2 \cdot (-1)} h^2) = n^{-3} h^2 \notin P_3.$$ Therefore, $P_3$ is not normal in $G$. The only other possibility admitted by Sylow is that there are seven $3$-Sylow groups.

Obviously, $P_2$ is normalized by $h$, and by $$n h^3 n^{-1} = n n^{2^3 \cdot (-1)} h^3 = n^{-7} h^3 = h^3\in P_2$$ it is normalized by $n$, too. Therefore, $P_2$ is normal in $\langle n,h\rangle = G$. Hence $P_2$ is the unique $2$-Sylow group.

azimut
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