Show that $\alpha = (1 2 3)(2 3 4)(5 6 7)(7 8 9 10)$ has order $10$ in $S_n$ $(n\geq10)$.

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So I have done problem #13 in section 7.5 of t book Abstract Algebra: An Intorduction, 3rd Edition by Thomas W. Hungerford (ISBN: 978-1-1115696-2-4). multiple times over now, but I still get the order of $\alpha$ is 6. Here is some of my work:

$\alpha = (12)(34)(5678910)$ $\alpha^2 = (579)(6810)$ $\alpha^3 = (12)(34)(58)(69)(710)$ $\alpha^6 = \alpha^3 • \alpha^3 = [(12)(34)(58)(69)(710)] • [(12)(34)(58)(69)(710)] = (1)$

Even by Theorem 7.25 I am getting that the order of $\alpha$ is 6. Theorem 7.25 states:

The order of a permutation $\tau$ in $S_n$ is the least common multiple of the lengths of the disjoint cycles whose product is $\tau$.

Is there something I am doing incorrectly, or is the book wrong?

P.S. I tried to post images of my work and the book question but “I need at least 10 reputation to post images” or so the error message said.

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You're right. The decomposition into disjoint cycles yields $$(1\, 2\, 3)(2\, 3 \,4)(5\, 6\, 7)(7 \,8\, 9\, 10)=(1\,2)(3\,4) (5\,6\,7\,8\,9\,10).$$