I know that I need to look for an element in $S_{10}$ with order 25 (because $\mathbb{Z_5}$ $\times$ $\mathbb{Z_5}$ is isomorphic to $\mathbb{Z_{25}}$?). Can I find an element with such an order in $S_{10}$?
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Z5 x Z5 is isomorphic with Z25: wrong. Precisely because the former has no element of order 25. + Please format your formulas. Hint: find two commuting elements of order $5$ in $S_{10}$. – Anne Bauval May 10 '24 at 09:07
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I actually was looking for the fact if Z5 x Z5 is isomorphic to Z25. I guess I checked wrong. Two cycles (1,2,3,4,5)(6,7,8,9,10) would be an element with order 5 right? – addae May 10 '24 at 09:16
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More precisely, these two $5$-cycles are two commuting elements of order 5, hence they generate a subgroup of $S_{10}$ isomorphic to $\Bbb Z_5\times\Bbb Z_5$. – Anne Bauval May 10 '24 at 09:20
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Note moreover that there is no element of order $25$ in $S_{10}$. See for instance this post or that one. The order of any element of $S_{10}$ is a divisor of $2^3\cdot3^2\cdot5\cdot7$. – Anne Bauval May 10 '24 at 09:35
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The minimal faithful permutation degree of $C_5\times C_5$ is $10$ - see the Theorem in this answer. So $S_{10}$ is the smallest symmetric group having a subgroup isomorphic to $C_5\times C_5$. – Dietrich Burde May 10 '24 at 17:13