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I'm asked to prove that the order of $H:=\langle(12345),(2354)\rangle$ is $20$, in an Algebra exercise 2nd year maths.

What I have achieved is that $20\mid |H|$ and $|H|=20,120$ since $H$ can't be the alternating group and there are no subgroups of order 40 in $S_5$. But I don't know how to prove $H\neq S_5$.

Thanks!

moqui
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1 Answers1

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Consider the integers modulo $5$. Let $\alpha$ be the operation $\alpha\colon x\mapsto x+1$. Let $\beta$ be the operation $\beta\colon x\mapsto 2x$.

Then $\alpha,\beta$ generate all affine automorphisms of $\mathbb{Z}/5\mathbb{Z}$, which has size $20$: $$x\mapsto ax+b,$$ with $a\in \{1,2,3,4\}$ and $b\in \{0,1,2,3,4,5\}$.


As pointed out by @JeanMarie, we have shifted the indices by $-1$ here, so $\alpha$ is the permutation $(01234)$ and $\beta$ is the permutation $(1243)$.

tkf
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    [+1] Very astute, but for an understanding of the isomorphism between the initial issue and yours, you should first explain that you shift everything by $-1$, working in $\mathbb{Z/5Z}$ with $(01234)$ and $(1243) \equiv (1248)$ – Jean Marie Jan 07 '22 at 11:21
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    A similar kind of trick was used here – Jean Marie Jan 07 '22 at 11:40
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    It seems this is a very nice answer. Unfortunately, I haven't studied automorphisms. – moqui Jan 07 '22 at 12:01
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    @moqui No problem, an automorphism is just a bijective group homomorphism. It is as common for group theory as prime numbers for number theory. – Dietrich Burde Jan 07 '22 at 12:17
  • @moqui Think of the maps $x\mapsto ax+b$ with $a\neq 0$ as a particular subset of permutations of ${0,1,2,3,4}$. You need to check that such functions are closed under composition and inverse - then you know they are a subgroup. There are $20$ such functions ($4$ choices for $a$ and $5$ choices for $b$). You also need to check that these are distinct as permutations ($ax+b=a'x+b'$ for all $x$ implies $a=a', b=b'$). Then you can conclude that $\alpha,\beta$ generate a subgroup of order $20$. With the shift of $-1$ in index, you can identify $\alpha,\beta$ with your two permutations. – tkf Jan 07 '22 at 18:52