Subnormal p-Sylow Subgroup of Finite Group

[A.Magnus]

I am self-studying a class note on finite group and come across a problem like this:

PROBLEM: Let $G$ be a dihedral group of order 30. Determine $O_2(G),O_3(G),O_5(G), E(G),F(G)$ and $R(G).$

Where $O_p(G)$ is the subgroup generated by all subnormal p-subgroups of $G$; $E(G)$ is the layer subgroup $G$; $F(G)$ is the fitting subgroup of $G$ and $R(G) := E(G)F(G)$ is the radical subgroup of $G.$

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Let's focus on determining $O_2(G),O_3(G),O_5(G)$ first, and here are what I got so far:

(1) Suppose that the number of $Syl_p(G)$ in a group is denoted by $n_p$, and the order $|Syl_p(G)| := o_p.$

(2) We observe that $|G| = 30 = 2 \cdot 3 \cdot 5, \$ all of which are prime, and $(2, 15) = 1,$ therefore if we take $p = 2$ then by Sylow's Theorem $n_2 \equiv 1 (mode \ 2)$ and $n_2 \mid 15.$ Therefore $n_2 = \{1, 3, 5, 15 \}.$ The theorem further implies that $o_2 = 2.$

(3) If we take $p = 3$, we observe that $(3, 10) = 1$ therefore by the same Sylow's Theorem we get $n_3 = \{ 1, 10 \}$ and $o_3 = 3.$

(4) Again if we take $p = 5$, we observe that $(5, 6) = 1$ therefore by the same Sylow's Theorem we get $n_5 = \{ 1, 6 \}$ and $o_6 = 6.$

(5) Observe that $Syl_2(G), Syl_3(G)$ and $Syl_5(G)$ are cyclic since their orders are prime, therefore their elements do not overlap, meaning that their intersection consists only of $\{ e \}$, the neutral element.

(6) Recall that $Syl_2(G), Syl_3(G)$ and $Syl_5(G)$ have to co-exist, and $|G| = 30,$ therefore the only possible combination will be $n_2 = \{1, 3, 5\}, n_3 = \{1\},$ and $n_5 = \{1\}.$ Otherwise the total elements $|G| > 30.$

(7) ...

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Unfortunately I don't know what is next after line (6), especially I don't know how to incorporate the fact that they are sub-normal and the group is dihedral. Nor do I know if I have been on the right direction. I would therefore love to get help from you. The class note does not offers any hint in solving the problem except a parade of lemmas, corollaries and theorems, one of which I believe is relevant to solving this problem:

Lemma: Let $p$ be a prime, then (i) $O_p(G)$ is the uniquely determined largest normal p-subgroup of $G$; (ii) $O_p(G)$ is the intersection of all Sylow p-subgroups of $G.$

Thank you very much for your time and help.

 

[jbunniii]

Maybe I'm missing something, but to me it seems like overkill to use the Sylow theorems to deduce the number of each type of Sylow subgroup. You are not dealing with some arbitrary group $G$ with $30$ elements; you're given that it's the dihedral group, which is easy to analyze.

In step (6), I'm not sure how you excluded $n_2 = 15$. In fact, $n_2$ does equal $15$. To see this, observe that the group is generated by two elements $r$ and $s$ satisfying $r^{15} = s^2 = 1$ and $rs = sr^{-1}$. Every element of the form $r^k s$ has order $2$, since $(r^k s)(r^k s) = (sr^{-k})(r^k s) = 1$, and there are $15$ elements of this form.

As for the other Sylow subgroups, their orders are $3$ and $5$, so they can't contain any of the $r^k s$ elements, which have order $2$. Therefore they can only contain elements of the form $r^k$, hence they are subgroups of $\langle r \rangle$. Since a finite cyclic group has exactly one subgroup with each possible order, $n_3 = n_5 = 1$ as you concluded.

 

[A.Magnus]

@jbunniii : I am so sorry I did not respond to you immediately, somehow I did not received any email alert. Perhaps the alert went to Greg Bernhardt instead. Here are my response to you:

(1) In hindsight, you are right and I was wrong on $n_2$, it should include $n_2 = 15.$ The final conclusion should be $n_2 = \{1, 3, 5, 15 \}.$ Note $n_p$ stands for the count of $Sly_p(G)$ in $G,$ and that we both agree $n_3 = n_5 = 1.$

(2) The reason I am using $Syl_p(G)$ is that I wanted to take advantage of the lemma found on the same class note: "The group $O_p(G)$ is the intersection of all Sylow p-Subgroups of $G$."

(2a) Thus, using the above little lemma, I can say that if $n_2 = 1,$ then $O_2(G) = \{a^0, a^1\}$ since we have only one $Syl_2(G).$ Notice that since $p = 2$ is prime, therefore $O_2(G)$ is cyclic. Now, knowing that $G$ is dihedral $D_{15}$, please refresh me with the basic knowledge: How do you determine the generator $a$?

(2b) If $n_2 = 3, 5,$ or $15,$ then $O_2(G) = \{1\}$ since they have the same identity. Correct me if I am wrong here.

(3) Using the same little lemma, I can say that $O_3(G)$ is the intersection of all $Syl_3(G)$'s. But since $n_3 = 1,$ there is only one $Syl_3(G),$ therefore $O_2(G) = \{b^0, b^1, b^2\}.$ Again, refresh me with finding out the generator $b.$

(4) The same with $O_5(G).$ Since $n_5 = 1$ therefore $O_5(G) = \{c^0, c^1, c^2, c^3, c^4 \}.$ I need to determine the generator $c.$

Please advise and thank you very much for your time and help.

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PS with a meta question: How do you indent to the right (or tab) the (2a) and (2b), such that readers will recognize that they are the sub-paragraph of (2)? Thanks again.

 

[jbunniii]

@A.Magnus - I'll respond point by point.

(1) You are correct that $n_2 \in \{1,3,5,15\}$. However, since we know that the group is the dihedral group of order 30, in fact $n_2 = 15$.

(2) Understood, you need to know $Syl_p(G)$ for $p=2,3,5$. But because we are dealing with a specific group of order 30 (the dihedral group), we don't need to use the Sylow theorems in order to determine $Syl_p(G)$. The analysis is quite elementary for this group, as I pointed out in my previous message. But certainly there is no harm in using the Sylow theorems if you are able to reach the same conclusion, namely $n_2 = 15$ and $n_3 = n_5 = 1$.

(2a/b) In fact $n_2 = 15$, so as you noted, $O_2(G) = \{1\}$ since any pair of distinct subgroups of prime order can only intersect trivially.

(3) I think you mean $O_3(G) = \{b^0, b^1, b^2\}$. Here, of course, $b^0 = 1$. This subgroup is generated by the element $r^5$ where, as I mentioned in the previous message, $r$ is the "rotation" and $s$ is the "flip" about one of the axes of symmetry.

(4) Yes, $O_5(G)$ is the unique subgroup of order 5. It is generated by $r^3$ where $r$ is again as above.

Not sure about the indentation question :D

 

[A.Magnus]

@jbunniii : I am very, very, very grateful to get connected with you, your patience and clarity of mind have helped me putting the first parts of the problem (on $O_2(G), O_3(G),$ and $O_5(G)$) into closure. It also came as a surprise because a friendly answer finally came only from an unexpected source -- the Physics Forum, as I have been posting the same question on Math Stack Exchange here for more then one week but got only lukewarm responses. Thank you again and again!

Now on the second parts of the problem, $F(G), E(G)$ and $R(G),$ here are what I got so far:

(1) My note defines $F(G)$, the fitting subgroup, simply as "the (complex) product of all subgroups of $O_p(G)$ with $p$ a prime number. So in this case, the solution is straightforward, I think, but correct me if I was wrong:

$\begin{align} F(G) &= O_2(G) O_3(G) O_5(G) \\ &= \{1\} \{1, r^5, r^10\} \{1, r^3, r^6, r^9, r^{12}\} \\ &= \{ 1, r^2, r^3 ... r^{13}, r^{14} \} \end{align}$
​

(2) On $E(G)$, my note gives a very long and winding chain of definition that is useless for my purpose. However, there is a little lemma that I think might give me a hand:

$[E(G), F(G)] = \{1\},$
​

where it is defined further as:

$\begin{align} [E(G), F(G)] &:= \langle [e, f] \mid e \in E(G), f \in F(G) \rangle, \\ [e, f] &:= e^{-1}f^{-1}ef \end{align}$​

Since $[e, f] = 1,$ does it mean that $e = f$ and therefore $E(G) = F(G)$? I would love to get your take in this one.

(3) Finally, my note defines the Radical of finite group $R(G)$ as $R(G) := E(G) F(G).$ If my analysis in (2) is correct, then it is doable to get $R(G)$ but it will be very tedious since $|E(G)| = |F(G)| = 15.$ Is there any shortcut?

Thanks again and again to jbunniii. In the meantime, for other heavyweights beside jbunniii, you are most welcome to pitch in. Thanks.

​

 

[jbunniii]

@A.Magnus - Again, replying point by point:

(1) I think your $F(G)$ is correct. The Fitting subgroup is by definition the largest normal nilpotent subgroup of $G$. Certainly $H = \{1, r, r^2, r^3, \ldots, r^{14}\}$ is a normal subgroup (since it has index 2) and it is nilpotent (since it is abelian). The only subgroup larger than $H$ is $G$ itself, but $G$ is not nilpotent.

(2) No, $[e,f] = 1$ does not imply $e=f$. By definition, $[e,f] = e^{-1}f^{-1}ef$ . This is called the commutator of $e$ and $f$. Note that $[e,f] = 1$ if and only if $e^{-1}f^{-1}ef = 1$ if and only if $ef = fe$, i.e., $e$ and $f$ commute.

I hadn't heard of the layer subgroup until now. But assuming your characterization is correct: $[E(G), F(G)] = \langle [e,f] | e \in E(G), f \in F(G)\rangle$ is the subgroup generated by the commutators of the form $e^{-1}f^{-1}ef$ where $e\in E(G)$ and $f \in F(G)$. If $[E(G),F(G)] = 1$ then certainly $e^{-1}f^{-1}ef = 1$ for all $e\in E(G)$ and $f \in F(G)$.

Recall from (1) that $F(G) = \{1, r^2, r^3, \ldots, r^{14}\} = \langle r\rangle$. Now what elements of $G$ commute with every element in $F(G)$? It's easy to see that the elements of $F(G) = \langle r \rangle$ commute with each other, since $F(G)$ is cyclic. It's also easy to verify that these are the only elements of $G$ that commute with every element of $F(G)$.

So, assuming your characterization is correct, namely that $[E(G),F(G)] = 1$, it follows that $E(G)$ must be a subgroup of $F(G)$. Do you have any additional information about the definition of the layer subgroup that would allow you to identify it more precisely?

(3) Assuming what I wrote in (2) is correct, namely that $E(G)$ is a subgroup of $F(G)$, it's easy to determine $E(G)F(G)$. Do you see why?

By the way, I would be curious to know the context of this problem. It seems (apologies if I am wrong) that you are working with some fairly sophisticated concepts in group theory without being completely comfortable with the basics.

 

[jbunniii]

It also came as a surprise because a friendly answer finally came only from an unexpected source -- the Physics Forum, as I have been posting the same question on Math Stack Exchange here for more then one week but got only lukewarm responses.

As you know, MSE users live and breathe for reputation points. I bet if you offer a 50 point bounty you would get some answers, probably better than mine. :D

 

[A.Magnus]

As you know, MSE users live and breathe for reputation points. I bet if you offer a 50 point bounty you would get some answers, probably better than mine. :D

If you define "better than mine" as answer that is slicker, more elegant and more poem-like, then I really do not need it. All I need is a line-by-line patient explanation just like yours. Thanks again.

 

[A.Magnus]

(1) I think your $F(G)$ is correct. The Fitting subgroup is by definition the largest normal nilpotent subgroup of $G$. Certainly $H={1,r,r^2,r^3,…,r^{14}}$ is a normal subgroup (since it has index 2) and it is nilpotent (since it is abelian). The only subgroup larger than $H$ is $G$ itself, but $G$ is not nilpotent.

Thank you! I got $F(G)$ now!

(2) No, $[e,f] = 1$ does not imply $e = f .$ By definition, $[e,f] = e^{-1}f^{-1}ef .$ This is called the commutator of $e$ and $f.$ Note that $[e,f] = 1$ if and only if $e^{-1}f^{-1}ef = 1$ if and only if $ef = fe,$ i.e., $e$ and $f$ commute.

I got it now, $[E(G), F(G)] = \{1\} \Rightarrow [e, f] = 1 \Rightarrow ef = fe,$ i.e., $e$ and $f$ commute.

So, assuming your characterization is correct, namely that $[E(G),F(G)] = 1,$ it follows that $E(G)$ must be a subgroup of $F(G).$ Do you have any additional information about the definition of the layer subgroup that would allow you to identify it more precisely?

Taking the clue from you, I rummaged the class note one more time and came up with this small item tucked and wedged among parades upon parades of lemmas, theorems and corollaries:
​

$E(G) \lhd G$​

Since $G = \{1, r^1, r^2, r^3, \ldots r^{14}, s^0, s^1, s^3 \ldots s^{14} \},$ and since $\forall e \in E(G), \forall g \in G, eg = ge,$ therefore the only possible solution would be $E(G) = \{1\}.$ This will nicely fit your claim that $E(G) \leq G$ and also the lemma $[E(G), F(G)] = \{1\}.$ Let me know if there is mistake on my reasoning.

By the way, I would be curious to know the context of this problem. It seems (apologies if I am wrong) that you are working with some fairly sophisticated concepts in group theory without being completely comfortable with the basics.

This problem is from a class note for this 2015 Spring semester's Group Theory class, which is set to have the first day of class next Monday 01/12, but I study on my own earlier while other students are still sleeping. The chapter that this problem is under is The Radical of a Finite Group. This is a graduate class, I am a HS math teacher and am taking it remotely online. This is only my second class on Abstract Algebra, perhaps for this reason the materials are very uphill for me but I enjoy it. (For your curiosity I don't mind sending you the class note but I need to look around for a third-party drop-off website so that I do not have to know your email address.)

Time permitting, I would love to publish the complete set of solution to this problem here and at MSE to benefit other learners. To be sure I will attribute all the credits to you: If this question's complete solution set is a bouquet, then all the stems and flowers are yours while mine is only the tiny string that ties them together. Thank you again and again for your time and extreme patience.

 

[A.Magnus]

As you know, MSE users live and breathe for reputation points. I bet if you offer a 50 point bounty you would get some answers, probably better than mine. :D

@Jbunnii : Do you mind if you take a look at this question here? I have been posting left and right for help but could not get any. Thanks for you time.