How to find $N_G(H) $?

Question

Consider the symmetric group $S_{15}$ and assume $G=S_{15}$ throughout this post. The subgroups $H, K\le G$ are said to be conjugate if $\exists \sigma\in G$ such that $H=\sigma K\sigma^{-1}$.

Let $H$ be a subgroup of order $15$ .Then find the number of subgroups conjugate to $H$.

Attempt: Consider the group action of $G$ on the class of all subgroups.

Then $\textrm{Orbit}(H) =\textrm{Conj}(H) $

$\textrm{Stab}(H) =N_G(H) $

By Orbit-stabilizer theorem, $|\textrm{Conj}(H) |=[G:N_G(H) ]$

$$N_G(H)=\{\sigma \in G: \sigma H\sigma^{-1}\subseteq H\}$$

More information:

1. $H\le G$ is cyclic.

2. $H$ can't be normal in $G$

3. $\sigma, \tau\in G$ are conjugate $\iff$ have same cycle type.

How to calculate $N_G(H) $ ?

Answer 1

The question is, in my opinion, a bit weak. Posting this also as a way for you to show a real attempt by doing the listed small exercises.

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As Chris pointed out in the comments, the calculation is different for different kinds of subgroups $H\le S_{15}$ of order fifteen. Let's first take a look at the various cases. Prove the following (may be done earlier in the course already):

1. The group $H$ is necessarily cyclic.
2. There exists an element of order five $\sigma\in H$ and an element $\tau\in H$ of order three. They must commute with each other.
3. Clearly $\sigma$ is a product of disjoint $5$-cycles, and $\tau$ is a product disjoint $3$-cycles. The different cases are determined by the structural relations of these two permutations. Show that if $\sigma$ has three disjoint $5$-cycles, then $\tau$ must similarly move all fifteen points (= a product of five disjoint $3$-cycles), and in this case $\sigma\tau$ is a $15$-cycle. Warning: This step is a bit trickier than the others. It may be a good idea to first convince yourself of the fact that in the group $S_5$ no $5$-cycle commutes with any $3$-cycle.
4. Show that in the other cases $\sigma$ and $\tau$ are disjoint. In other words, $\sigma$ has one or two disjoint $5$-cycles, and $\tau$ has, space permitting, one or two or three disjoint $3$-cycles.

Ok. At this point you know a bit more about the possible ways a group of order fifteen can sit inside $S_{15}$. As you will see, this allows us to describe the elements of the normalizer. I will do the simple case where $\sigma$ is a (single) $5$-cycle, and $\tau$ is a disjoint $3$-cycle.

• Why can we assume without loss of generality that $\sigma=(12345)$ and $\tau=(678)$?
• If $\alpha$ is an element of the normalizer $N=N_{S_{15}}(H)$ why does this imply that $\alpha\sigma\alpha^{-1}$ must be a power of $\sigma$ and that $\alpha\tau\alpha^{-1}$ must be a power of $\tau$?
• Why does the previous bullet imply that $\alpha$ must permute the numbers in the subset $A=\{1,2,3,4,5\}$ among each other, and similarly permute the numbers in the subset $B=\{6,7,8\}$ within this subset? Observe that as a consequence of this $\alpha$ must also permute the numbers of the complementary subset $C=\{9,10,11,12,13,14,15\}$.
• Ok, so in light of the previous bullet it follows that the restriction of $\alpha$ to the subset $A$ must be an element of the normalizer of the subgroup $\langle\sigma\rangle\le Sym(A)$, and the restriction of $\alpha$ to the subset $B$ must be an element of the normalizer of the subgroup $\langle\tau\rangle\le Sym(B)$.
• Let $N_1$ be the normalizer of $\langle(12345)\rangle$ in $S_5$, and let $N_2$ be the normalizer of $\langle(123)\rangle$ in $S_3$. Calculate the respective orders of $N_1$ and $N_2$ using what you know about the smaller groups $S_5$ (resp. $S_3$) and the normalizers of Sylow $5$-subgroups (resp. Sylow $3$-subgroups) in there.
• Why is $N$ isomorphic to a direct product $N_1\times N_2\times S_7$?

Ok, so that gives you the answer in this case. The other cases are different. Particularly in the second bullet the elements of the normalizer may also permute the disjoint cycles present in $\sigma$ or $\tau$. You need to pay attention to such details. For example:

• The elements $(12345)(6789A)$ (here $A$ is a stand-in symbol replacing $10$) and $(16)(27)(38)49)(5A)$ commute.
• The elements $(123)(456)(789)$ and $(147)(258)(369)$ also commute.

Good luck!

Answer 2

Several facts are useful:

(i) A group of order $15$ is cyclic and we have $\phi(15)=8$ choices for generators.

(ii) Two elements in $S_n$ are conjugate if and only if their ${\it{types}}$ coincide. Here, we say that $\alpha\in S_n$ has type $1^{p}2^q3^r\cdots$ if $\alpha$, expressed in a product of disjoint cycles, is a product of $p$ cycles of length 1, $q$ cycles of length 2, $r$ cycles of length 3, and so on. (We note that length 1 cycles correspond to fixed points. They are usually omitted.)

For example, in $S_{15}$, $$\alpha=(1234567890abcde), \beta=(12345)(67890)(abc), \gamma=(12345)(678)(90a)(bcd), \delta=(12345)(678)(90a), \epsilon=(12345)(678)$$ have types $15^1$, $1^2 3^1 5^2$, $1^13^35^1$, $1^43^25^1$, $1^73^15^1$ respectively. Here we used $0:=10, a:=11,\dots,e:=15$ as symbols on which $S_{15}$ acts.

(iii) For $\sigma\in S_n$ of type $1^{p_1}2^{p_2}3^{p_3}\cdots$, the order $e$ of $\sigma$ is the LCM of $\{i\mid p_i>0\}$. Moreover, for $d$ prime to $e$, $\sigma^d$ has the same type as $\sigma$. These assertions are also easy to see.

With these preparations, we can compute the conjugacies of subgroups of order 15 inside $S_{15}$. By (i), every $H<G$ of order 15 is cyclic with a generator $\sigma$. In fact, possible types of $\sigma\in S_{15}$ are covered by the five examples $\alpha,\dots,\epsilon$ above, by using (iii). It also shows that the type of $\sigma$ is independent of the choice of $\sigma$. Therefore, by (ii), two subgroups $\langle \sigma \rangle$ and $\langle \tau \rangle$ are conjugate in $S_{15}$ if and only if $\sigma,\tau$ has the same types (since any conjugacy is given by a conjugacy of suitably chosen generators). It follows that there are exactly five conjugacy classes of order 15 subgroups (represented by $\alpha,\dots,\epsilon$ above). To count the number of conjugate subgroups in each case, we just need to count the number of elements with that type and mod out by $8=\phi(15)$ by (i), corresponding to the choice of generators. These numbers are computed as follows: we choose the set of symbols appearing in each cycle, then count the number of possible cycles, then take quotients by possible symmetry. $$\begin{split} \text{Case }\alpha&\colon \frac{\begin{pmatrix}15\\15\end{pmatrix}14!}{8}=10897286400\\ \text{Case }\beta&\colon \frac{\begin{pmatrix}15\\5\end{pmatrix}4!\cdot \begin{pmatrix}10\\5\end{pmatrix}4!\cdot \begin{pmatrix}5\\3\end{pmatrix}2! }{8\cdot 2!}=544864320\\ \text{Case }\gamma&\colon \frac{\begin{pmatrix}15\\5\end{pmatrix}4!\cdot\begin{pmatrix}10\\3\end{pmatrix}2!\cdot\begin{pmatrix}7\\3\end{pmatrix}2!\cdot\begin{pmatrix}4\\3\end{pmatrix}2!}{8\cdot 3!}=201801600\\ \text{Case }\delta &\colon\frac{\begin{pmatrix}15\\5\end{pmatrix}4!\cdot\begin{pmatrix}10\\3\end{pmatrix}2!\cdot\begin{pmatrix}7\\3\end{pmatrix}2!}{8\cdot 2!}=75675600\\ \text{Case }\epsilon&\colon\frac{\begin{pmatrix}15\\5\end{pmatrix}4!\cdot\begin{pmatrix}10\\3\end{pmatrix}2!}{8}=2162160 \end{split}$$ With these numbers in hand, we can also compute the order of normalizers as $$\begin{split} \# N_G(\langle \alpha \rangle)&= 120\\ \# N_G(\langle \beta \rangle)&=2400\\ \# N_G(\langle \gamma \rangle)&=6480\\ \# N_G(\langle \delta \rangle)&=17280\\ \# N_G(\langle \epsilon \rangle)&=604800 \end{split}$$

Edit: It would be worth clarifying the structure of normalizers. After the computation of their orders, we are allowed to just guess and confirm the possible normalizers, and then prove the equality by the coincidence of orders. We denote a cyclic group of order $d$ by $C_d$ for simplicity. It is useful to know that for a $C_{15}=\langle g\rangle$, $\mathrm{Aut}(C_{15})$ is generated by the two elements $$g\mapsto g^7 \text{ and }g\mapsto g^{11}.$$

Cases $\alpha$ and $\epsilon$ are the simplest, since we don't have cycles of duplicate lengths inside. $$\begin{split} \text{Case }\alpha\colon N_G(\langle \alpha\rangle)&=\langle \alpha,\ (285d)(3e9b)(47c0),\ (2b)(38)(5e)(6a)(9d)\rangle\\ &\simeq C_{15}\rtimes \mathrm{Aut}(C_{15}).\\ \text{Case }\epsilon\colon N_G(\langle \epsilon\rangle)&=\langle \epsilon,\ (2354),\ (78),\ \text{(permutations fixing $1,\dots,8$)}\rangle\\ &\simeq (C_{15}\rtimes \mathrm{Aut}(C_{15}))\times S_{7}. \end{split}$$ Cases $\beta,\gamma,\delta$ needs more care, since the normalizer can mix up cycles of identical lengths, as already pointed in the answer by Jyrki. In fact, we can choose the representatives of such elements in the centralizer subgroup $C_G$. $$\begin{split} \text{Case }\beta\colon C_G(\langle \beta\rangle)&=\langle (12345),\ (67890),\ (abc),\ (16)(27)(38)(49)(50),\ (de)\rangle\\ &\simeq (C_5\wr S_2)\times C_3\times S_2\ \text{ (of order 300)}\\ N_G(\langle \beta\rangle)&=\langle C_G(\langle \beta\rangle),\ (2354)(7809),\ (bc)\rangle\\ &\simeq C_G(\langle \beta\rangle)\rtimes \mathrm{Aut}(C_{15})\\ \text{Case }\gamma\colon C_G(\langle \gamma\rangle)&=\langle (12345),\ (678),\ (90a),\ (bcd),\ (69b)(70c)(8ad),\ (69)(70)(8a)\rangle\\ &\simeq C_5\times (C_3\wr S_3)\ \text{ (of order 810)}\\ N_G(\langle \gamma\rangle)&=\langle C_G(\langle \gamma\rangle),\ (2354),\ (78)(0a)(cd)\rangle\\ &\simeq C_G(\langle \gamma\rangle)\rtimes \mathrm{Aut}(C_{15})\\ \text{Case }\delta\colon C_G(\langle \delta\rangle)&=\langle (12345),\ (678),\ (90a),\ (69)(70)(8a),\ \text{(permutations fixing $1,\dots,a$)}\rangle\\ &\simeq C_5\times (C_3\wr S_2)\times S_4\ \text{ (of order 2160)}\\ N_G(\langle \delta\rangle)&=\langle C_G(\langle \delta\rangle),\ (2354),\ (78)(0a)\rangle\\ &\simeq C_G(\langle \delta\rangle)\rtimes \mathrm{Aut}(C_{15}) \end{split}$$

Edit 2: Our computation so far may be generalized into the following statement. Let $\alpha$ be an element in $S_n$, whose type is $1^{a_1}2^{a_2}3^{a_3}\cdots$, where $\sum ia_i=n.$ Then, the normalizer $N_{S_n}(\langle \alpha\rangle)$ is isomorphic to a group $$\bigl(\prod_i (C_i\wr S_{a_i})\bigr)\rtimes \mathrm{Aut}(C_{l}) $$ where $l$ is the order of $\alpha$.　In particular, the normalizer has order $$(\prod_i i^{a_i} (a_i)!)\cdot \phi(l). $$