Problem 5.2.9 (Subgroups orf order $2$ or index $2$ in $ Z_{60} imes Z_{45} imes Z_{12} imes Z_{36}$)

Question

Let $A = Z_{60} 	imes Z_{45} 	imes Z_{12} 	imes Z_{36}$. Find the number of elements of order $2$ and the number of subgroups of index $2$ in $A$.

Richard Ganaye · Accepted Answer

\begin{proof} The decomposition of $A$ in elementary divisors is given by
\begin{align*}
A&= Z_{60} 	imes Z_{45} 	imes Z_{12} 	imes Z_{36}\
&\simeq (Z_{2^2} 	imes Z_{3} 	imes Z_{5}) 	imes (Z_{3^2} 	imes Z_5)	imes (Z_{2^2} 	imes Z_{3}) 	imes (Z_{2^2} 	imes Z_{3^2})\
&\simeq (Z_{2^2} 	imes Z_{2^2} 	imes Z_{2^2})  	imes (Z_{3^2}	imes Z_{3^2}	imes Z_{3}	imes Z_{3})	imes (Z_5	imes Z_5).
\end{align*}
By exercise 7 and 8, since $2$ divides $|A|$,  the number $N$ of elements of order $2$ is equal to the number of subgroups of index $2$ in $A$. It is given by
$$N = 2^t - 1 =2^3 - 1 = 7.$$
(where $t$ is the number of elementary divisors of $A$ relative to the prime factor $2$).
\end{proof}

Problem 5.2.9 (Subgroups orf order $2$ or index $2$ in $ Z_{60} \times Z_{45} \times Z_{12} \times Z_{36}$)

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Problem 5.2.9 (Subgroups orf order $2$ or index $2$ in $ Z_{60} \times Z_{45} \times Z_{12} \times Z_{36}$)

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