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Let $G$ be a group of order $|G| = 126 = 2 \cdot 3^2 \cdot 7$.

Let $H \leq G$ be a subgroup of order $|H| = 14$ and $\varphi:G \rightarrow H$ be a surjective group homomorphism.

How many 3-Sylow groups are in $G$?

(Let $s_3$ be the number of 3-Sylow groups)

My try

According to Sylow theorems, I know that:

  • $s_3 | 14 \Rightarrow s_3 \in \{1, 2, 7, 14\}$
  • $s_3 = 2 \cdot r + 1 \Rightarrow s_3 \in \{1, 7\}$

But now I'm stuck. How can I make use of $\varphi$?

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    $\begingroup$ What do you know of $\ker \varphi$? $\endgroup$ Sep 2, 2013 at 15:20
  • $\begingroup$ What is the order of $\ker(\varphi)$? What do you know about normal Sylow subgroups? $\endgroup$ Sep 2, 2013 at 15:20
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    $\begingroup$ I know that $\varphi$ is not injective as $14 = |H| < |G| < \infty$. So $\ker \varphi$ is not trivial. And I know that 2 p-Sylow groups are conjugate to each other. $\endgroup$ Sep 2, 2013 at 15:22
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    $\begingroup$ How are the orders of $G,\, H$ and $\ker \varphi$ related? $\endgroup$ Sep 2, 2013 at 15:24
  • $\begingroup$ As $\ker \varphi$ is a subgroup of $G$ I know (becaus of Lagrange): $\ker \varphi | \#G = 126$. Ah, and $\frac{\#G}{\# \ker \varphi} = \# H$ because of homomorphism theorem. So $\ker \varphi = 9$ $\endgroup$ Sep 2, 2013 at 15:32

1 Answer 1

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Hints: The kernel of $\varphi$ is a normal subgroup of $G,$ and all $3$-Sylow subgroups of $G$ are conjugate. Since $\varphi:G\to H$ is a surjective homomorphism with $|G|=126$ and $|H|=14,$ what is $|\ker\varphi|$?

The only shaky thing about this proof is: how do we know that there is a surjective homomorphism $\varphi:G\to H?$

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  • $\begingroup$ $|\ker \varphi| = 9$. So I know that $\ker \varphi$ is a 3-sylow group. But as long as I don't have a second one, I don't see how this helps. I know that there is such an surjective homomorphism, because it is given (this is a task in an old exam). $\endgroup$ Sep 2, 2013 at 15:36
  • $\begingroup$ Excellent! Since $\ker\varphi$ is a normal subgroup of $G$, what can you say of its conjugates? Since all $3$-Sylow subgroups of $G$ are conjugate to $\ker\varphi$, then what does that allow us to conclude? $\endgroup$ Sep 2, 2013 at 15:38
  • $\begingroup$ Normal subgroups are invariant under conjugation. So the answer is: $s_3 = 1$. Thank you very much for leading me to the answer! $\endgroup$ Sep 2, 2013 at 16:12

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