1
$\begingroup$

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!

$\endgroup$
3
  • 1
    $\begingroup$ Can't you just calculate that $(2354)^{-1}(12345)(2354)=(13524)=(12345)^2$? $\endgroup$ Jan 7 at 10:58
  • 1
    $\begingroup$ What would this imply? $\endgroup$
    – moqui
    Jan 7 at 11:10
  • 2
    $\begingroup$ 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 $20$.] $\endgroup$ Jan 7 at 13:04

1 Answer 1

5
$\begingroup$

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 $(1243)$.

$\endgroup$
5
  • 3
    $\begingroup$ [+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 $(1243) \equiv (1248)$ $\endgroup$
    – Jean Marie
    Jan 7 at 11:21
  • 1
    $\begingroup$ A similar kind of trick was used here $\endgroup$
    – Jean Marie
    Jan 7 at 11:40
  • 2
    $\begingroup$ It seems this is a very nice answer. Unfortunately, I haven't studied automorphisms. $\endgroup$
    – moqui
    Jan 7 at 12:01
  • 2
    $\begingroup$ @moqui No problem, an automorphism is just a bijective group homomorphism. It is as common for group theory as prime numbers for number theory. $\endgroup$ Jan 7 at 12:17
  • $\begingroup$ @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. $\endgroup$
    – tkf
    Jan 7 at 18:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.