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Let $\sigma=(13579)(24)(68)$ and $\tau=(15937)(26)(48)$ be elements in $S_9$. Prove that $$\langle \sigma,\tau\rangle\cong \mathbb{Z}/5\mathbb{Z}\times V_4.$$

In a previous question, I showed that $\sigma^2=\tau^6$. I know that, in order to find an isomorphism, I need to either find a bijective homomorphism or to make use of some theorem in my book which says that if $H_1,H_2$ are subgroups of $G$ that commute, have intersection $\{e\}$ and where every $g\in G$ can be written as $g=h_1 h_2$ with $h_1\in H_1$, $h_2\in H_2$, then $G\cong H_1\times H_2$. However the subgroups $\langle\sigma\rangle$ as well as $\langle\tau\rangle$ have order 10, and $\mathbb{Z}/5\mathbb{Z}$ has order 5 and $V_4$ order 4.

Could someone provide a hint, I know this should be easy.

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  • $\begingroup$ What group is $V_4$? $\endgroup$ – G Tony Jacobs Mar 26 '18 at 12:46
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    $\begingroup$ @GTonyJacobs The Klein four group. $\endgroup$ – anon Mar 26 '18 at 12:46
  • $\begingroup$ Please add a link to the previous question. $\endgroup$ – lhf Mar 26 '18 at 13:53
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If a permutation is a product of $5$-cycles and $2$-cycles, then there are powers you can take it to to extract out the $5$-cycles and $2$-cycles. For instance,

$$ \sigma^5=(24)(68). $$

Use this to show that $\langle\sigma,\tau\rangle=\langle (24)(68),(26)(48),(13579),(15937)\rangle$. Note the first two permutations generate a subgroup of $\mathrm{Perm}(\{2,4,6,8\})$ and the other permutations each generate a subgroup of $\mathrm{Perm}(\{1,3,5,7,9\})$; the result will be an internal direct product.

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  • $\begingroup$ I don't understand the first three lines. How did you get that $\langle\sigma,\tau\rangle=\langle(24)(68),(26)(48),(13579),(15937)\rangle$? $\endgroup$ – Heinz Doofenschmirtz Mar 26 '18 at 13:03
  • $\begingroup$ If $g=ab$ in a group, where $a$ and $b$ commute, then $g^n=a^nb^n$. In particular, if a permutation is a product of disjoint cycles, then those disjoint cycles commute, and taking the permutation to a power amounts to taking each cycle to that power. Since $\sigma$ is a product of a $5$-cycle and two $2$-cycles, taking it to the $5$th power will kill off the $5$-cycle and, well, taking $2$-cycles to odd powers doesn't change them, so this successfully siphons out the $2$-cycles from $\sigma$. I'll let you figure out how to get everything else in the $\langle\cdots\rangle$ I've given. $\endgroup$ – anon Mar 26 '18 at 13:05
  • $\begingroup$ Note another fact I'm using here: if $A\subseteq\langle B\rangle$ (if you can generate every element of $A$ using stuff from $B$) and $B\subseteq\langle A\rangle$ (every element of $B$ can be generated using stuff from $A$) then we must have $\langle A\rangle=\langle B\rangle$. $\endgroup$ – anon Mar 26 '18 at 13:06
  • $\begingroup$ I showed that $\langle (24)(68),(26)(48)\rangle\cong V_4$ and that $\langle(13579),(15937)\rangle=\langle (13579)\rangle\cong \mathbf{Z}/5\mathbf{Z}$. Since the cycles are disjoint, they commute, the intersection is $\{e\}$ and the third condition for my lemma I didn't yet confirm. Is this correct? $\endgroup$ – Heinz Doofenschmirtz Mar 26 '18 at 13:21
  • $\begingroup$ @HeinzDoofenschmirtz Correct. $\endgroup$ – anon Mar 26 '18 at 13:33

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