Let $G$ be a group of order $5\times 13\times 43\times 73$. Find the number of elements of order $5$.

Here is what I do:

Since $|G| = 5m$ where $(m,5) = 1$, $m = 13\times 43\times 73$, by Sylow's Theorem, there exists a Sylow $5$-subgroup with #Sylow $5$-subgroup dividing $13\times 43\times 73$ and #Sylow $5$-subgroup = $5k + 1$ for some $k \geq 0.$ So I think that the only possible choice is $1$. This yields that there is only one Sylow $5$-subgroup which yields that the all $4$ elements of this cyclic group except the identity is of order $5$. So there are only $4$ elements of order $5$ in this group.

Is my approach to this question correct?

  • 2
    $\begingroup$ Yes, that works. Notice that the divisors of $13\cdot 43\cdot 73$ are $\equiv 3$ or $9$ or $7\mod 10$, so cannot be $\equiv 1 \mod 5$. $\endgroup$ – user138530 Dec 12 '15 at 2:56

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