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I am struggling to solve this. According to $(123)$, $1$ gets sent to $2$, but according to $(456)(1457)$, $1$ gets sent to $5$. Can this be decomposed?

Shaun
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    Read this https://math.stackexchange.com/questions/31763/multiplication-in-permutation-groups-written-in-cyclic-notation – B.Swan Oct 25 '19 at 21:30
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    You are right; $(456)(1457)$ sends $1$ to $5$. Now $(1,2,3)$ leaves $5$ fixed, so ... – saulspatz Oct 25 '19 at 21:36

1 Answers1

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$$1\stackrel{(1\,4\,5\,7)}\longmapsto 4 \stackrel{(4\,5\,6)}\longmapsto 5\stackrel{(1\,2\,3)}\longmapsto 5$$ $$5\stackrel{(1\,4\,5\,7)}\longmapsto 7 \stackrel{(4\,5\,6)}\longmapsto 7\stackrel{(1\,2\,3)}\longmapsto 7$$ $$7\stackrel{(1\,4\,5\,7)}\longmapsto 1 \stackrel{(4\,5\,6)}\longmapsto 1\stackrel{(1\,2\,3)}\longmapsto 2$$ $$2\stackrel{(1\,4\,5\,7)}\longmapsto 2 \stackrel{(4\,5\,6)}\longmapsto 2\stackrel{(1\,2\,3)}\longmapsto 3$$ $$3\stackrel{(1\,4\,5\,7)}\longmapsto 3 \stackrel{(4\,5\,6)}\longmapsto 3\stackrel{(1\,2\,3)}\longmapsto 1$$ so that one of the cycles is $$(1\,5\,7\,2\,3). $$ $$4\stackrel{(1\,4\,5\,7)}\longmapsto 5 \stackrel{(4\,5\,6)}\longmapsto 6\stackrel{(1\,2\,3)}\longmapsto 6$$ $$6\stackrel{(1\,4\,5\,7)}\longmapsto 6 \stackrel{(4\,5\,6)}\longmapsto 4\stackrel{(1\,2\,3)}\longmapsto 4$$ so that another cycles is $$(4\,6) $$ and there cannot be more.

Hagen von Eitzen
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