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

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 $G.$

The text does not offers much help in solving the problem except a parade of lemmas, corollaries and theorems, one of them I believe is relevant:

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 $G.$

I would love to hear any helps, hints or links to solve the problem. If you have a choice of between a slick-elegant solution versus a down-to-earth dummy one, do please give the latter - you have a turtle not a rabbit over here :-) Thank you for all your times and help.

POST SCRIPT - 1: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am not giving up with this problem yet! Let's focus on determining $O_2(G),O_3(G),O_5(G)$ first, hopefully from them I will get hints for solving the next parts of question. Here are my attempts so far:

(1) Observe the above LEMMA (ii) that says $O_p(G)$ is the intersection of all Sylow p-subgroups of $G,$ meaning that $O_2(G), O_3(G)$, and $O_5(G)$ are the intersections of subgroups of respectively $Syl_2(G), Syl_3(G)$ and $Syl_5(G)$.

(2) Suppose that the number of $Syl_p(G)$ in $G$ is denoted by $n_p$, and the order of $|Syl_p(G)|:=o_p.$ I can provide computation leading to conclusion that $n_2=\{1,3,5\}$, $n_3=\{1\}$, $n_5=\{1\}$ and $o_2 = 2, o_3 = 3, o_5 = 5.$ All are cyclic since the orders are primes.

(3) ...

Unfortunately I don't know what is next after line (2), nor do I know if I have been on the right direction. I would therefore love to get help from you. Thank you very much for your time and help.

POST SCRIPT - 2: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
See complete solution below through helps from "jbunniii" of Physics Forum. Thanks to all who have been helping me.

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  • $\begingroup$ So, let's start with $O_2$ then? What do Sylow 2-subgroups look like? What's their intersection? $\endgroup$ – Myself Dec 27 '14 at 0:08
  • $\begingroup$ Hints: It should be stated somewhere that $F(G)$ is the (unique) largest nilpotent normal subgroup of $G$. Your dihedral group has a very large nilpotent normal subgroup Also, $E(G)$ is not solvable when $E(G) \neq 1$ (for any group $G$). $\endgroup$ – Geoff Robinson Dec 27 '14 at 1:56
  • $\begingroup$ More people would see your question if you gave it the "group-theory" tag. $\endgroup$ – Derek Holt Dec 28 '14 at 17:02
  • $\begingroup$ @DerekHolt : Just add Group Theory tag, thanks for reminding me. $\endgroup$ – Amanda.M Dec 28 '14 at 23:30
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After days of research, I am very happy to finally be able to put together a solution for this question. But to all fairness, I did not do it all by myself -- I got tremendous line-by-line helps from "jbunniii" at Physics Forum here. If this solution is a bouquet, all stems and flowers should go to "jbunniii" while I claim only the small tiny string that binds them together. However, any mistakes herein should be blamed on me alone.

For solution to this question, I will rely on the following lemmas from my class notes:

(a) The group $O_p(G)$ is the intersection of all Sylow p-subgroups of $G$;
(b) $[E(G), F(G)] = \{1\}$, where $E(G)$ and $F(G)$ are respectively the Layer and Fitting of $G$;
(c) $E(G) \lhd G$.

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(A) FINDING $O_2(G):$
(1) Suppose that the number of $Syl_p(G)$ subgroup in $G$ is denoted by $n_p$ and the order of subgroup $Syl_p(G)$ by $o_p$;
(2) Here $|G| = 30 = 2 \cdot 3 \cdot 5$, implying that $G$ is a dihedral of 15-gon $D_{15}$;
(3) Per Sylow's Theorems, $n_2 \equiv 1 \pmod 2, n_2 \mid 15$ and $o_2 = 2.$ Therefore $n_2 = \{1, 3, 5, 15 \}$;
(4) Recall that $G = D_{15} = \{r^0, r^1, r^2 \ldots r^{14}, s^0, s^1, s^2 \ldots s^{14} \}$ where $r$ and $s$ stand for rotation and reflection respectively. Also observe that $r^{15} = s^2 = 1$ and $rs = sr^{-1}$;
(5) Since $o_2 = 2$ therefore $Syl_2(G)$ is cyclic and its elements must have order of 2;
(6) Of all elements of $D_{15}$, we observe that for $k = \{ 1, 2, \ldots 15 \}$, $r^ks$ is the only element that has order of 2, this is because $(r^ks)^2 = (r^ks)(r^ks) = (sr^{-k})(r^ks) = s^2 = 1$;
(7) Therefore of $n_2 = \{1, 3, 5, 15\}$, it has to be $n_2 = 15$ only because $k = \{ 1, 2, \ldots 15 \}$;
(8) Thus per Lemma (a) above, $O_2(G) = \bigcap _{i = 1}^{15} (Syl_2(G))_i = \{1\}$

(B) FINDING $O_3(G):$
(1) Using the same analysis, we get $n_3 = \{1, 10 \}$. However, $n_3 = 10$ is not possible since $o_3 = 3$, and $Syl_2(G), Syl_3(G)$ and $Syl_5(G)$ have to co-exist as subgroups of $G$ and $|G|= 30$. Therefore the only choice is for $n_3 = 1$;
(2) Of all elements of $D_{15}$, only $r^5$ has order 3 since $(r^5)^3 = r^{15} = 1;$
(3) Thus per Lemma (a) above, $O_3(G) = Syl_3(G) = \{1, r^5, r^{10}\}$.

(C) FINDING $O_5(G):$
(1) Again using the same analysis, we get $n_5 = \{1, 6 \}$, and the only choice is for $n_5 = 1$;
(2) Of all elements of $D_{15}$, only $r^3$ has order 5 since $(r^3)^5 = r^{15} = 1;$
(3) Thus per Lemma (a) above, $O_5(G) = Syl_5(G) = \{1, r^3, r^6,r^9,r^{12}\}$.

(D) FINDING $F(G):$
Per definition from my note, $F(G)$, the fitting subgroup, is the (complex) product of all subgroups of $O_p(G)$ with $p$ a prime number:
$$\begin{align*} F(G)&=O_2(G)O_3(G)O_5(G) \\ &=\{1\}\{1,r^5,r^{10}\}\{1,r^3,r^6,r^9,r^{12}\} \\ &=\{1,r^2,r^3...r^{13},r^{14}\} \end{align*}$$

(E) FINDING $E(G):$
We are going to use the Lemma (c) above that $E(G)\lhd G$.​ Since $G = \{1,r^1, r^2, r^3,\dots r^{14},s^0, s^1, s^3 \ldots s^{14} \}$, and since $\forall e \in E(G),\forall g \in G,eg=ge$, therefore the only possible solution would be $E(G)=\{1\}$. Observe that this fits nicely the Lemma (b) that $[E(G),F(G)]=\{1\}$.

(F) FINDING $R(G):$
Per definition from my note again, the radical of $G$: $$\begin{align*} R(G) :&= F(G)E(G) \\ &=\{1,r^2,r^3...r^{13},r^{14}\} \{1\} \\ &=\{1,r^2,r^3...r^{13},r^{14}\} \\ \end{align*} $$
$\blacksquare$

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