Sylow subgroup of group of order 56

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May 29th 2010, 11:45 AM
Giraffro

Sylow subgroup of group of order 56

Let $G$ be a group of order 56, $T \in \text{Syl}_2(G), S \in \text{Syl}_7(G)$ be given. Then $|T| = 8, |S| = 7$ . Assume $S$ is not normal, so then $T \lhd G$ . Show that $T \cong C_2^3$ .

Here's what I have so far. My idea is to show that each element of $T$ has order dividing 2. Let $z \in S$ be given such that $\langle z \rangle = S$ . Then $G / T = \{T z^i : i=0,\dots,6\}$ and $\forall x \in T, (Tz)^x = Tz$ . I've also got that $G = T \rtimes S$ , although I fail to seen how any of this is helping.

Any advice?
May 29th 2010, 02:28 PM
tonio

Quote:

Originally Posted by Giraffro

Let $G$ be a group of order 56, $T \in \text{Syl}_2(G), S \in \text{Syl}_7(G)$ be given. Then $|T| = 8, |S| = 7$ . Assume $S$ is not normal, so then $T \lhd G$ . Show that $T \cong C_2^3$ .

Here's what I have so far. My idea is to show that each element of $T$ has order dividing 2. Let $z \in S$ be given such that $\langle z \rangle = S$ . Then $G / T = \{T z^i : i=0,\dots,6\}$ and $\forall x \in T, (Tz)^x = Tz$ . I've also got that $G = T \rtimes S$ , although I fail to seen how any of this is helping.

Any advice?

So you have that $T\rtimes S\Longrightarrow$ there exists a non-trivial homomorphism $S\rightarrow Aut(T)$ . Well, what are the possible orders of automorphisms groups of groups of order 8?

Tonio
May 29th 2010, 04:29 PM
Giraffro

Quote:

Originally Posted by tonio

So you have that $T\rtimes S\Longrightarrow$ there exists a non-trivial homomorphism $S\rightarrow Aut(T)$ . Well, what are the possible orders of automorphisms groups of groups of order 8?

Tonio

So $\exists \phi : S \to \text{Aut}(T)$ a group homomorphism such that $\forall x, y \in T, \forall i, j \in \mathbb{Z}, x z^i y z^j = x \phi(z^i)(y) z^{i+j}$ . If $\phi(z) = \text{id}_T$ then $\forall x \in T, z^x = x^{-1} z x = x^{-1} \phi(z)(x) z = z$ , so as $S = \langle z \rangle, S \lhd G$ , which is a contradiction. Therefore $\phi(z) \ne \text{id}_T$ and so as $o(\phi(z)) \ne o(z), o(\phi(z)) = 7$ . So it follows that $\exists x \in T$ such that $x,\phi(z)(x),\dots,\phi(z^6)(x)$ are distinct and have the same order. So $T$ has at least 7 elements of a specific order.

$ \begin{array}{c|ccc} & \text{Order} \, 2 & \text{Order} \, 4 & \text{Order} \, 8 \\ \hline C_8 & 1 & 2 & 4 \\ C_4 \times C_2 & 3 & 4 & 0 \\ C_2^3 & 7 & 0 & 0 \\ D_8 & 5 & 2 & 0 \\ Q & 1 & 6 & 0 \end{array}$

Therefore $T \cong C_2^3$ .

Thanks for the help!