Finding the number of elements with order 2 in a group of given order.

Question

Let $A=Z_{60} \times Z_{45} \times Z_{12} \times Z_{36}$. Find the number of elements of order 2.

My proof so far:

By factoring A, we get that
\begin{align*} A \cong Z_4 \times Z_4 \times Z_4 \times B \end{align*} Where $B$ is a direct product of groups of odd order and therefore does not contain elements of order 2, and can therefore be ignored. Now let $G=Z_{4}^{'} \times Z_{4}^{''} \times Z_{4}^{'''}$, and let $a \in Z_{4}^{'}$, $b \in Z_{4}^{''}$ and $c \in Z_{4}^{'''}$, with $|a|=|b|=|c|=2$. Since there can only be one element of order 2 in $Z_4$ $a,b,c$ are the only elements with this property. Then the following combinations would grant an element in $A$ with order 2. \begin{align*} (a,1,1),(a,b,1),(a,b,c),(1,b,1),(1,b,c),(a,1,c),(1,1,c) \end{align*} Which would grant 7 element in $A$ with order 2.

Thanks for the help.

Your factorization of the group is not correct. Note that $Z_4$ is not isomorphic to $Z_2\times Z_2$. — Tobias Kildetoft, Mar 18 '14 at 13:25
If $G=A\times B$, with $a\in A$ and $b\in B$ having order $2$, then $(a,1)$, $(1,b)$ and $(a,b)$ have order $2$. What for a product of three groups? — egreg, Mar 18 '14 at 13:28
The revised proof is correct, although you should extend $(a,1,1)$ to be $(a,1,1,1)$ to include a component from $B$ (and similarly for the other elements). — vadim123, Mar 18 '14 at 14:04

score 1 · Answer 1 · answered Mar 18 '14 at 14:15

1

Hint: If $G=A\times B$, with $a\in A$ and $b\in B$ having order $2$, then $(a,1)$, $(1,b)$ and $(a,b)$ have order $2$. Conversely, if $(x,y)$ has order $2$, then $(x^2,y^2)=1$, so…

What for a product of three groups?

answered Mar 18 '14 at 14:15

egreg

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Finding the number of elements with order 2 in a group of given order.

1 Answers1