# Action on a normal subgroup where each coset acts freely

Given a group G and a normal subgroup N of G, is there an action of G on N such that, whenever g,h are distinct members of the same N-coset, we have g•n≠h•n? If not, then can this be done in the case G is abelian?

Take for granted that we may select a set of coset representatives for the N-cosets to use as "origins" for each N-coset. I'm working on some abstract analysis/descriptive set theory ideas, and got stuck on this thought because the algebra got a little too far out of my element. If it can't be done, a counterexample would be GREATLY appreciated.

Okay. Gotta update this question: In this setting, the groups are all either countably infinite or of size continuum.

Also it's important to actually be able to describe the action.

• @FrançoisBrunault I think it can, as $g,h$ are supposed to lie in the same coset - finding two distinct elements in the same coset when $N=1$ seems difficult... – Dirk Oct 6 '17 at 8:41
• You surely thought about it for some time. Could you maybe give an example of an action that has the desired property in a non-trivial case (i.e. $N \neq 1$ and $N \neq G$)? I'm having a hard time coming up with an operation that leaves $N$ fixed as a set but every coset "acts" regularly on $N$... – Dirk Oct 6 '17 at 8:46
• Isn't the cyclic group $C_4$ of order 4 with $|N|=2$ a counterexample? There is only one nontrivial action, and that does not have the specified property. – Derek Holt Oct 6 '17 at 9:25
• In the case $G=\mathbf{Z}/4\mathbf{Z}$ and $N=\{0,2\}$, by your assumption the element $2 \in N$ must act as a transposition of $N$. Then the action of $1$ on $N$ is a permutation $\sigma$ which satisfies $\sigma^2 = (0 2)$, which is impossible. – François Brunault Oct 6 '17 at 9:28
• "An action of $G$ on $N$": how is the action related to the group structure of $N$? of its embedding in $G$? I understand the question as equivalent to "is there an action on $G$ a set $X$ of the same cardinality as $N$ that is free on $N$"? If $N$ is finite, this is equivalent to require that $N$ is part of a semidirect decomposition. But if both $N,G$ are infinite countable, this is always true. – YCor Oct 6 '17 at 16:56

• To have an action of a group $G$ the ingredients are a set $X$ and a homomorphism from $G$ into $S_X.$ The action is faithful if for all $g_1$, aside from the identity, there is an $x$ with $g_1\cdot x \neq x.$ If $X \subset X'$ then this is also an action on $X'$ which fixes the things outside $X.$ It remains faithful if it already was. Of course if $X$ has some mathematical structure it may help define the homomorphism more explicitly. If there is a subgroup $H$ with an appropriate subset $Y$ of the same cardinality as $X$ then we can transfer the action of $G$ on $X$ to an action of $G$ on $Y.$ Then the action becomes an action on $H$ (and indeed all of $G$) which happens to leave unmoved everything in $G$ (and $H$) which is not in $Y.$ The condition which you seem to want for $Y$ to be appropriate is that there is a constructive bijection $f$ from $X$ to $Y.$ Then we can define $g\cdot y=f(g\cdot (f^{-1}y)).$

• Your cosets condition can be reduced to saying that the action of $G$ on $X$ (which might be $Y$ or $H$) ,when restricted to $H$, is faithful. If $G$ itself acts faithfully on $X$ this is automatic.

I call the subgroup $H$ because the condition that it be normal in $G$ need not be important. In fact it could just be a subset $Y$ which is large enough. In the first two examples I will use $G=S_{\mathbb{N}},$ the uncountable group of permutations of the positive integers. The action(s) of $G$ in these first two will also satisfy the stronger condition of being faithful: for any $g_1 \neq g_2$ in $G$ (rather than $g_1,g_2$ in a common coset of $H$) there is an $h$ with $g_1 \cdot h \neq g_2\cdot h.$

My first example will use a normal subgroup and a familiar action depending on it being normal. Let $H$ be the subgroup of permutations which fix all but finitely many integers (which is countable and normal). Note the important (to this construction) property that the center of $H$ is trivial. Then the action $g \cdot h=ghg^{-1}$ is a natural group action of $G$ on $H.$ What you desire is that $a$ and $ah_1$ never act in the same way on $H$ when $h_1$ is not the identity. (As mentioned above, it is enough to have $a$ the identity.) Pick an $h_2 \in H$ such that $h_1h_2 \neq h_2h_1$ and observe that $a\cdot h_2 \neq (ah_1) \cdot h_2.$ This construction works any time you have a group $G$ with a normal subgroup $H$ whose center is trivial. In this particular case the stronger property condition is satisfied because the center of the entire group is trivial so the action by conjugation is faithful.

The second construction is quite arbitrary and the action not that natural. I will give it in generality and then get more specific. I need some infinite subgroup $H$ along with a countable subset $Y$ (which could be all of $H$ but need not be.) At this point only $Y$ matters, the rest of $H$ doesn't. I do require that there be an enumeration $y_1,y_2,\cdots$ of the elements of $Y.$ The enumeration should be constructive, but can be as simple or complex as you like. Then just define this action of $G$ on $H$: $g\cdot h=h$ when $h \notin Y.$ However $g \cdot y_i=y_j$ where the permutation $g$ sends the integer $i$ to the integer $j.$ In other words each integer $i$ is replaced by the element $y_i \in H.$

This construction works any time you have a group $G$ acting faithfully on a set $X$ and a subgroup $H$ of cardinality greater than or equal to that of $X.$

Let me get more specific. I'll stick to cyclic groups so let $\sigma\in G$ have infinite order and let $H=<\sigma>$ be the countable cyclic subgroup it generates. I'll discuss some unusual choices of $\sigma$ at the end, but the construction is indifferent.

For the enumeration of a countable subset of $H$ I will take $y_1=\sigma^3,y_2=\sigma^5,y_3=\sigma^8,y_4=\sigma^{13},\cdots$ i.e. the powers of $\sigma$ whose exponents are Fibonacci numbers greater than $2,$ in order of increasing size. (Other, weirder, constructive orders could also be given)

There is no reason not to take $Y$ to be all of $H,$ but also no reason to do so.

There are obvious nice choices for $\sigma$ but it works just as well to take

$\sigma=(\cdots 29\ 17\ 13\ 5\ 3\ 7\ 11\ 19 \cdots)$ which moves only the odd primes or

$\sigma=(10\ 11)(100\ 101\ 102)(1000\ 1001 \ 1002\ 1003)\cdots$ which has cycles of each order and moves only numbers of the form $10^i+j$ with $1 \le i$ and $0 \leq j \leq i.$

LATER The third construction, since you request it, is one with $G=\mathbb{R}$ and $H=\mathbb{Z}$ and the operation is addition, however the field structure is used. I will start with an uncountable subgroup $J$ with countable index in $\mathbb{R}$. Then you grant in the question that there is a usable enumerated set of coset representatives $X=\{x_1=0,x_2,x_3,\cdots\}$ so that each real $g$ is uniquely $g=x_i+j$ for a $j \in J.$ The action is $g \cdot x_s=x_t$ where $x_t$ is the coset representative of $g+x_s$ (and of $x_i+x_s.$) I then will take the subset $Y$ to be the positive integers in $\mathbb{Z}$ (though I think I have established that weirder choices are equally usable.) And, as before, define an action of $\mathbb{R}$ on $Y$ (and on $\mathbb{Z}$) by $g\cdot i=j$ exactly when $g\cdot x_i=x_j$ in the previous action.

It remains only to describe $J.$ Maybe there are better choices or constructions of a $J,$ but this occurs to me. It will have the property that $X=\mathbb{Q}$ is a set of coset representatives so enumeration is easy. The rationals $\mathbb{Q}$ are a sub-field of $\mathbb{R}$ which is thus a $\mathbb{Q}$-vector space. I'll take (as you seem to allow) that there is a specific uncountable set of coset representative $S$ (with $1 \in S$) so that each real is uniquely a linear combination $z=q_0+q_1s_{i_1}+q_2s_{i+2}+\cdots +q_ms_{i_m}$ where $m=m_z$ is finite and the $q_i$ are non-zero except that $q_0$ is allowed to be $0$. ( i.e. $S$ is a Hamel basis .) My $J$ is precisely those $z \in \mathbb{R}$ which do have $q_0=0.$ This is a subgroup and the set of coset representatives , as mentioned, can be taken to be $X=\mathbb{Q}.$

• The problem is "arbitrarily identify". I need to be able to describe or define what's going on. – Michael Cotton Oct 7 '17 at 0:43
• Also, what's a representation? I'm not an algebraist. – Michael Cotton Oct 7 '17 at 0:59
• Now that I know you want infinite groups I will rewrite the answer. All I meant by arbitrarily identify was "pick a constructive bijection, any one you want" – Aaron Meyerowitz Oct 7 '17 at 6:40
• Can something similar be done on something more like $\mathbb{R}$ and the subgroup $\mathbb{Z}$, or are we leaning very hard on the structure of the permutation group here? – Michael Cotton Oct 7 '17 at 17:20
• I'm not sure in your descriptive set theory work what you accept as "constructive." SInce you grant that a subgroup always has a usable set of coset representatives, it might be pretty general. At any rate, I have given a last example which seems to me to work, modulo being allowed a basis for $\mathbb{R}$ as a $\mathbb{Q}$-vector space (i.e. a set of coset representatives). – Aaron Meyerowitz Oct 8 '17 at 1:03