# Determine order of $\langle(12345),(2354)\rangle$.

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!

• Can't you just calculate that $(2354)^{-1}(12345)(2354)=(13524)=(12345)^2$? Jan 7 at 10:58
• What would this imply? Jan 7 at 11:10
• That $H$ has a normal cyclic subgroup of order $5$ and hence $H=\langle (12345)\rangle \langle (2354)\rangle$ has order 20 [and is (as the solution of @tkf also shows) the Frobenius group of order $20$.] Jan 7 at 13:04

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 $$(1243)$$.

• [+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 $(1243) \equiv (1248)$ Jan 7 at 11:21
• A similar kind of trick was used here Jan 7 at 11:40
• It seems this is a very nice answer. Unfortunately, I haven't studied automorphisms. Jan 7 at 12:01
• @moqui No problem, an automorphism is just a bijective group homomorphism. It is as common for group theory as prime numbers for number theory. Jan 7 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 7 at 18:52