If $\langle a\rangle$ is a cyclic group of order $90$, write $a$ as the product of $p$-elements, for various primes $p$.

Question

We say that $a$ is a $p$-element if the order of $a$ is $p^n$ for some integer $n\ge0$. The primes dividing $p$ are $2,3$ and $5$.

Thus the $2$ - elements are $\{e,a^{45}\}$.

The $3$ - elements are $\{e,a^{10},a^{20},a^{30},a^{40},a^{50},a^{60},a^{70},a^{80},a^{90}\}$.

The $5$ - elements are $\{e,a^{18},a^{36},a^{54},a^{72}\}$

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How can I write $a$ as product of these?

Answer 1

Since $a^{36},a^{10} $ and $a^{45} $ have orders $5,9$ and $2$, respectively, you can take their product.