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Consider the group $Q_{8}$ and $S_{3}$. How many elements of the group $Q_{8}\times S_{_{3}}$ have degree 12?

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What have you tried? –  Tobias Kildetoft Jan 23 '13 at 18:27
    
If an element of $Q_8 \times S_3$ is written $(g,x)$ for $g \in Q_8$ and $x \in S_3$, what are the conditions on the orders of $g$ and $x$ in their respective groups for $(g,x)$ to have order 12? –  orlandpm Jan 23 '13 at 18:30
    
@orlandpm Something does not seem to mind –  aliakbar Jan 23 '13 at 18:31
    
In the notation I used before, do you know what $(g,x)\cdot (h,y)$ is? –  orlandpm Jan 23 '13 at 18:37
    
@orlandpm yes. SO the order of $(g,x)$ become $gcd(o(g),o(x))$ –  aliakbar Jan 23 '13 at 18:39

1 Answer 1

up vote 2 down vote accepted

If $(g,x)$ is an element of $Q_8 \times S_3$, its order is the least common multiple of the orders of $g$ in $Q_8$ and $x$ in $S_3$.

The group $Q_8$ has one element of order $2$ and six elements of order $4$. The group $S_3$ has two elements of order $3$ and three element of order $2$.

I'll leave it to you to count pairs of elements whose orders have $LCM =12$.

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