# Direct product

• January 7th 2009, 03:19 AM
aliceinwonderland
Commutator subgroup
Let $G_{1} \times G_{2}=Z_{12} \times S_{24}$.
1. The commutator subgroup of $Z_{12} \times S_{24}$ I got is $e \times A_{24}$.
2. The quotient group of $\frac{Z_{12} \times S_{24}}{e \times A_{24}} = Z_{12} \times Z_{2}$.

Is above 1&2 correct?
Since the maximum order of elements in $Z_{12} \times Z_{2}$ is 12, the above 2 seems wrong. ((12 * 24!) / (0.5 *24!) = 24 ).
If 2 is wrong, what is the quotient group of 2?
• January 7th 2009, 07:52 AM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
Let $G_{1} \times G_{2}=Z_{12} \times S_{24}$.
1. The commutator subgroup of $Z_{12} \times S_{24}$ I got is $e \times A_{24}$.
2. The quotient group of $\frac{Z_{12} \times S_{24}}{e \times A_{24}} = Z_{12} \times Z_{2}$.

Is above 1&2 correct?
Since the maximum order of elements in $Z_{12} \times Z_{2}$ is 12, the above 2 seems wrong. ((12 * 24!) / (0.5 *24!) = 24 ).
If 2 is wrong, what is the quotient group of 2?

We need to know that $(G_1\times G_2)' = G_1' \times G_2'$. To prove this just consider a commutator, $c=(a_1,a_2)(b_1,b_2)(a_1^{-1},a_2^{-1})(b_1^{-1},b_2^{-1})$ where $a_1,b_1\in G_1$ and $a_2,b_2\in G_2$. We can rewrite $c = (a_1b_1a_1^{-1}b_1^{-1}, a_2b_2a_2^{-1}b_2^{-1})$. But $a_1b_1a_1^{-1}b_1^{-1}\in G_1'$ and $a_2b_2a_2^{-1}b_2^{-1} \in G_2'$. Therefore, the subgroup generated by the commutators i.e. $(G_1\times G_2)'$ is equal to $G_1'\times G_2'$.

Now the commutator subgroup of $\mathbb{Z}_{12}$ is $\{ e \}$ (not $e$ !) While it is an established result that the commutator of $S_{24}$ is $A_{24}$.
Therefore, I agree with your result.
• January 7th 2009, 08:41 PM
bulls6x
2. is definitely correct. And note that the order of (Z/12)X(Z/2) is 24 not 12. The highest order of a single element is 12 but there are 24 distinct elements in the whole group.