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Let's consider $G$ a group of order $|G| = 6 = 2 ·3 $. Therefore, using Sylow's Theorem we can find $n_3 = 1$ and $n_2>1$ or $n_2 = 3$ meaning there exists only one 3-Sylow subgroup, which is normal to G, and at least one 2-Sylow subgroup.

Consider $P$ is the normal subgroup of order 3 and $Q$ a subgroup of order 2. Now from Lagrange we know that the intersection of these subgroups is $P\cap Q = \{e\}$. Since $P\triangleleft G$ we know that $PQ = QP \iff PQ \leq G $. Now due to cardinality, being $|PQ| = \frac{|P||Q|}{|P\cap Q|} = 6 = |G|$ and $PQ \leq G $ we have $PQ = G$.

Since both $P$ and $Q$ have a prime order we have that $P \cong \mathbb{Z}/3\mathbb{Z}$ and $Q \cong \mathbb{Z}/2\mathbb{Z}$ so $ G = PQ \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. Now both of these groups are abelian, hence their direct product is also abelian.

I would conclude now that $G$ is abelian. However, it is very easy to see that $S_3$, group of order 6 is not abelian.

I am not able to figure out where my mistake is and it would be of great help if someone could give me a hint.

Thanks!

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  • $\begingroup$ $G=PQ$ does not imply $G=P\times Q$ $\endgroup$ – Hagen von Eitzen Apr 16 '17 at 18:32
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The mistake is when you say $$G = PQ \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z};$$ this is true if both $P$ and $Q$ are normal, but this is not necessary the case.

In fact, if only $P$ is normal, then $Q$ acts on $P$ by conjugation, and then you can say that $$G = PQ \cong \mathbb{Z}/3\mathbb{Z} \rtimes \mathbb{Z}/2\mathbb{Z}.$$ Now there are two possible actions of $\mathbb{Z}/2\mathbb{Z}$ on $\mathbb{Z}/3\mathbb{Z}$ : the trivial one and the one that sends every element to its inverse. The trivial action gives you the direct product $\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/6\mathbb{Z}$, the other one gives you the dihedral group $D_6\cong S_3$.

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Here $PQ=G$ as sets, but this does not imply that $P\times Q\cong G$ as groups. That only would be the case if every element of $P$ commutes with every element of $G$. In $S_3$ that isn't so; no element of order $2$ commutes with any element of order $3$ there.

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