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Let $G$ be a group and $H$ a subgroup of $G$ and let $[G:H]=n$. We consider here the question of whether there is an element in $g\in G$ such that $\left\{H,gH,\ldots,g^{n-1}H\right\}$ are all the cosets of $H$ in $G$.

(a) Show that if $n$ is not a prime this may fail.

(b) Show that if $n$ is prime such $g$ always exists. (Suggestion: Show first that a transtive subgroup of $S_n$ has order divisible by $n$. Show then that a transitive subgroup of $S_p$ has an element of order $p$. Use the coset representation to finish the proof. You may use Cauchy'theorem : a finite subgroup whose order is divisible by a prime $p$ has an element of order $p$.)

In (a), I can't find a simple counterexample.

In (b) by If $H$ is a subgroup of $G$ of prime index $p$, $\exists g\in G$ such that $G/H=\{H,gH,\ldots,g^{p-1}H\}$.

the solution is:

By considering the (transitive) action of $G$ on $G/H$, we have a homomorphism $$\psi:G\longrightarrow S_p$$ and thus $\psi(G)$ is a transitive subgroup of $S_p$. It follows that $p$ divides $\psi(G)$ and thus, by Cauchy's theorem, there is an element $\psi(g)\in\psi(G)$ of order $p$.

With the element $\psi(g)$ of order $p$, let us first verify that $g \notin H$. Suppose you had $g\in H$. Then $\psi(g^k)$ would fix $H$ for all $k$, so $\psi(g)$ would be a permutation of the $p-1$ other cosets of $H$, of order $p$. But $p \nmid (p-1)!$, so that can't be.

Hence we have $\psi(g)H \neq H$. Now consider the order of the orbit of $H$ under $\psi(g)^k$. Let that order be $n$. We saw $n > 1$. But $\psi(g)^p = \operatorname{id}$, hence $n \mid p$, so $n = p$, since $p$ is prime.

I don't see that:

(i) so $\psi(g)$ would be a permutation of the $p-1$ other cosets of $H$, of order $p$. But $p \nmid (p-1)!$, so that can't be.

My doubt: $\psi(g)$ would be a permutation of the $p-1$ other cosets of $H$, of order $p$. Then $p \mid (p-1)!$?

eraldcoil
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    For a counterexample when $n$ is not prime, let $G$ be the quaternion group $Q_8$ and $H = {1, -1}$. –  May 30 '20 at 05:21
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    Regarding your question about (i): could you be more specific about what you don't understand? If we go through the proof presented, then what is the first point at which something becomes unclear? – Ben Grossmann May 30 '20 at 05:21
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    Example for a: consider $G = \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$, and take $H = {(0,0,0),(1,0,0)}$. Here, $\Bbb Z_2 = \Bbb Z/2\Bbb Z$, which is to say that it is the cyclic group of order $2$. – Ben Grossmann May 30 '20 at 05:21
  • A permutation of $p-1$ elements can be identified with an element of $S_{p-1}$, which has order $(p-1)!$. Therefore such a permutation cannot have order $p$, since $p$ is prime and therefore does not divide $(p-1)!$. –  May 30 '20 at 05:33

2 Answers2

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For $a)$, consider $\Bbb Z\times\Bbb Z/(2\Bbb Z\times2\Bbb Z)$.

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Hint for part (a): Look for a counterexample with $H = \{1\}$.

For (b), if $\psi(g)$ fixes some element $k \in \{1, \dot, p\}$, then it belongs to the subgroup $K$ of $S_p$ consisting of those permutations fixing $k$. But $|K| = (p-1)!$. Since $p$ doesn't divide $(p-1)!$, this contradicts the fact that $\psi(g)$ has order $p$.

Incidentally, there's what I would consider an easier way to finish (b) by noting that any element of $S_p$ of order $p$ must be a $p$-cycle. Knowing that $\psi(g)$ acts as a $p$-cycle on the conjugates of $H$ tells you what you want to prove.

Anonymous
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  • I have that $G:=Z_2\times Z_2 and H=\left{(0,0)\right}$ works. $G/H=\left{H,(1,0)+H,(0,1)+H,(1,1)+H\right}$ and $(1,0)+(1,0)=(0,0), (0,1)+(0,1)=(0,0), (1,1)=(1,0)+(0,1)$ – eraldcoil May 30 '20 at 05:39
  • Your choice of $G$ and $H$ will work, but I don't understand what point you're making by saying $(1,1) = (1,0) + (0,1)$. If $H = {1}$, any non-cyclic $G$ will do. – Anonymous May 30 '20 at 05:42