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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 $\varphi$?

Martin Thoma
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1 Answers1

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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 $\varphi:G\to H?$

Cameron Buie
  • 102,994
  • $|\ker \varphi| = 9$. 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). – Martin Thoma Sep 02 '13 at 15:36
  • 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? – Cameron Buie Sep 02 '13 at 15:38
  • Normal subgroups are invariant under conjugation. So the answer is: $s_3 = 1$. Thank you very much for leading me to the answer! – Martin Thoma Sep 02 '13 at 16:12