Group action G on itself

**Bernhard** · October 21st 2011, 02:41 AM

Dummit and Foote define a group action as follows:

A group action of a group G on a set A is a map from G x A to A (written as g $\cdot$ a, for all g $\in$ G and a $\in$ A) satisfying the following properties

(1) $g_1$ $\cdot$ ( $g_2$ $\cdot$ a) = ( $g_1$ $g_2$ ) $\cdot$ a for all $g_1$ , $g_2$ $\in$ G and all a $\in$ A
(2) 1 $\cdot$ a = a for all a $\in$ A

Dummit and Foote Section 1.7 Group Actions Exercise 14 reads as follows:

Let G be a group and let A = G. SHow that if G is non-abelian then the maps defined by g $\cdot$ a = ag for all g, a $\in$ G do not satisy the axioms of a (left) group action of G on itself

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I am unsure of whether the operation of the group G on the set G - that is g $\cdot$ a can be treated the same as the operation of two elements of G as a group - that is ga.

Can anyone clarify this? [I suspect it cannot!]

Anyway, proceeding withthe problem we need to show

(1) $g_1$ $\cdot$ ( $g_2$ $\cdot$ a) = ( $g_1$ $g_2$ ) $\cdot$ a for all $g_1$ , $g_2$ $\in$ G and all a $\in$ A is NOT satisfied

Then $g_1$ $\cdot$ ( $g_2$ $\cdot$ a) = $g_1$ $\cdot$ (a $g_2$ )

The term on the right above is strictly a group action operation of $g_1$ on a set element a $g_2$

How do I progress this?

I do not think I can write ( $g_1$ a) $g_2$ as there is a group action mixed with a group operation ???

Am I overthinking this?

Peter

**emakarov** · October 21st 2011, 03:19 AM

Originally Posted by Bernhard

I am unsure of whether the operation of the group G on the set G - that is g $\cdot$ a can be treated the same as the operation of two elements of G as a group - that is ga.

I don't know if I parse this sentence correctly, but $g\cdot a=ag$ . On the left is the group action we are defining, and on the right it's the group operation on two group elements $a$ and $g$ .

Originally Posted by Bernhard

(1) $g_1$ $\cdot$ ( $g_2$ $\cdot$ a) = ( $g_1$ $g_2$ ) $\cdot$ a for all $g_1$ , $g_2$ $\in$ G and all a $\in$ A is NOT satisfied

Then $g_1$ $\cdot$ ( $g_2$ $\cdot$ a) = $g_1$ $\cdot$ (a $g_2$ )

It seems like you are starting with the claim you need to prove and then try to deduce something from it, since you say "then." This is incorrect.

$\begin{aligned}g_1\cdot (g_2\cdot a)&=g_1\cdot (ag_2)&&\mbox{by definition of action}\\&=(ag_2)g_1&&\mbox{by definition of action}\\&=a(g_2g_1)&&\mbox{by associativity of the group operation}\\&\ne a(g_1g_2)&&\mbox{in general, since the group is noncommutative}\\&=(g_1g_2)\cdot a&&\mbox{by definition of action}\end{aligned}$

**Bernhard** · October 21st 2011, 03:25 AM

Yes, of course, you are quite right - expressed it badly (wrongly)

Thanks for the help - see it clearly now!

Thanks

**Deveno** · October 21st 2011, 05:24 AM

G can act on itself by left-multiplication, by defining g.a = ga (because a, the "set element", is also an element of G).

the point of this exercise is to show that right multiplication is not a left action (it is a right action, though right actions are not often studied, being just the "mirror image" of left actions).

another way G can act on itself, is through conjugation: g.a = gag^-1.

there is yet another action of G that is important: suppose H is any subgroup of G (not necessarily a normal one).

we can form the set of left cosets {aH} of G, and define g.(aH) = (ga)H.

why would we do such a thing? well, the left regular action of G on G (g.a = ga) is an isomorphism of G with a subgroup of $S_{|G|}$ .

but G has |G| elements, and $S_{|G|}$ has (|G|)! elements, so G is perhaps just an itty bitty subgroup of a really large permutation group.

for example we can identify $\mathbb{Z}_4$ with the subgroup of S4 generated by (1 2 3 4). but $\mathbb{Z}_4$

has just 4 elements, whereas S4 has 24, so our cyclic group is "lost in the crowd".

on the other hand, if H is a large enough subgroup of G, then the set of left cosets might be pretty small, and if we're lucky, we might have an isomorphism of G into $S_{|G|/|H|}$ .

for example, D4 has 8 elements, but S8 is HUGE, it has 40,320 elements. so realizing D4 as a subgroup of S8, isn't very helpful.

but, we have the subgroup H = {1,s} of D4,with the following cosets $\{H, rH, r^2H, r^3H\}$ .

using this action, we can identify r with (1 2 3 4), $r \cdot (r^kH) = r^{k+1}H$ sends:

$H \to rH$
$rH \to r^2H$
$r^2H \to r^3H$
$r^3H \to H$

which is the same permutation as (1 2 3 4) just replacing the cosets by a number.

now, it's pretty clear that every power of r gives a different permutation, so the kernel of this action has order ≤ 2. what permutation does s correspond to?

well s.(H) = sH = H.and s.(r^2H) = (sr^2)H = {sr^2,sr^2s} = {sr^2, r^2} = r^2H. but:

s.(rH) = (sr)H = {sr,srs} = {r^3s, r^3} = r^3H, so s is not the identity, it's the permutation (1 3).

r --> (1 2 3 4)
r^2 --> (1 3)(2 4)
r^3 --> (1 4 3 2)
1 --> e
s --> ( 1 3)

so the image of D4 in S4 has at least 5 distinct elements, which means the kernel of the action has fewer than 2, so we must actually have an isomorphism of D4 into S4.

and this means that we have another way of looking at D4, as a subgroup of S4. which means that everything we know about permutation groups, we can apply to D4, we can talk about cycle lengths, and evenness/oddness. seeing D4 as "geometric" symmetries of the square, translates into D4 being a subgroup of the symmetric group S4 (sort of explains the name, huh?).

**Bernhard** · October 21st 2011, 03:40 PM

Thanks Deveno

I found that a REALLY helpful and informative post!

Peter

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