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I am studying Sylow theorems at the moment, more specifically trying to solve the following problem that I recently posted:

Let G be a finite group which has exactly eight Sylow 7 subgroups. Show that there exits a normal subgroup N of G such that the index [G:N] is divisible by 56 but not by 49

The link is here: If we have exactly 1 eight Sylow 7 subgroups, Show that there exits a normal subgroup $N$ of $G$ s.t. the index $[G:N]$ is divisible by 56 but not 49.

My initial response was to use Sylow's theorems to understand the order of the group. Though I am still very new at this. I was then told to look at this problem through group actions, which I know considerably less in. After taking a look at group actions again, I know just a few more things. Considering that I am just beginning at understanding this theory, I was wondering if someone can explain to me in laymen terms, the application of group actions and how to use them to solve problems. Of course, I know that I could just look up book definition, but I would like a little more insight than that. How would you explain group actions to someone with just calculus level knowledge maybe? What is the meaning of group actions, using language that is easy for a beginner to understand?

The definition of group action that I have found comes from Hungerford's Algebra. An action of a group $G$ on a set $S$ is a function $GxS \to S$ such that for all $x \in S$ and $g_1, g_2 \in G$: $ex=x$ and $(g_1g_2)x=g_1(g_2x)$

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  • $\begingroup$ What is "1 eight Sylow 7 subgroups" in blue? Please clarify. $\endgroup$ – P Vanchinathan Apr 6 '15 at 14:06
  • $\begingroup$ @PVanchinathan From the body of the linked problem, I think the "1" is spurious and it should just read "eight $p$-Sylow subgroups, where $p=7$". $\endgroup$ – Mario Carneiro Apr 6 '15 at 14:08
  • $\begingroup$ the question was a link to the problem but I have changed it to be more clear $\endgroup$ – cele Apr 6 '15 at 14:10
  • $\begingroup$ @cele tried to fix your formatting since you seemed to be having trouble with the link $\endgroup$ – Mario Carneiro Apr 6 '15 at 14:14
  • $\begingroup$ An interesting group action to consider as a start is action by conjugation. With this action, stabilisers of an element are those that commute with it, and elements in their own lone orbit (conjugacy class in this case) are in the centre of the group. Considering this action might help you get more of a grasp, as it did for me. $\endgroup$ – Myridium Apr 6 '15 at 14:17
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Note that another way to understand group actions is by "currying" the action into a function $G\to(S\to S)$; in this language a group action is just a homomorphism from $G$ to the permutation group on $S$. Thus in many ways the theory of group actions does not add anything more than you will get from understanding permutation groups and homomorphisms. Nonetheless, it can sometimes be useful to view group actions the "usual" way as a function $G\times S\to S$.

One great way to motivate the usage of group actions is to read the proof of Sylow's theorems. The "fundamental theorem" of group actions is the orbit-stabilizer theorem, and it is this that leads to most of the divisibility constraints in Sylow's theorems.

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A group action on a finite set of k elements is nothing but a homomorphism from the group to $S_{k}$. Now, since all the 7 sylow subgroups are conjugate, then G acts by conjugation on them. This gives rise to a homomorphism from $G$ to $S_{8}$. Let K be the kernel, then by the first isomorphism theorem $[G:K]$ divides 8!. Since 49 does not divide 8!, then 49 does not divides [G:K]. For the other part, you might want to use the transitivity of the action and the orbit stabilizer theorem to show that $[G:N]$ is divisible by 8.

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    $\begingroup$ Also, the divisibility of $|G:K|$ by $7$ comes from the fact that no Sylow $p$-subgroup can normalize a different Sylow $p$-subgroup, so the image of a Sylow $7$-subgroup under the homomorphism cannot be trivial. $\endgroup$ – Derek Holt Apr 6 '15 at 15:03
  • $\begingroup$ @DerekHolt, so essentially, by the orbit stablizer theorem, I have to show that $[G:N]=\frac{|G|}{|N|}$, and then I must show that N can divide 56 but not 49? $\endgroup$ – cele Apr 6 '15 at 19:22
  • $\begingroup$ @mich95, can you explain to me where 8! comes from? $\endgroup$ – cele Apr 6 '15 at 19:47
  • $\begingroup$ k! is the order of the symmetric group on k elements. $\endgroup$ – mich95 Apr 6 '15 at 19:48
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A group $G$ acting on a set $X$ means each element of $G$ leads to a permutation of the set. The effect of a permutation corresponding to $g\in G$ on an element $x\in X$ is written as $g.x$ or simply as $gx$. We can compose permutations and also apply group law on two elements of $G$. The permutations associated to these elements have to be such that an 'associative property' holds: that is what your last sentence means. and id element of the group should correspond to identity permutation. Another way of saying the same this that we have a group homomorphism from $G$ to the group of permutations of the set $X$. Note that two different elements of $G$ may lead to the same permutation.

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  • $\begingroup$ though this does help me understand the definition better, I am still struggling to see how this helps me solve the problem I have linked above. How do they correlate? $\endgroup$ – cele Apr 6 '15 at 14:22
  • $\begingroup$ Doesn't the answer by mich95 help to clarify things? $\endgroup$ – Derek Holt Apr 6 '15 at 17:00

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