As the title says: how can I find the subgroups of $S_n$? I know that $S_n = &lt;(1 2), (1 2 .. n)&gt;$ and $S_n = &lt;(1 i): i = 2, 3, .. n&gt;$. But how do I find all subgroups? And how do I know their orders?
95135
I am implementing an algorithm which finds every subgroup of given group.  Here's my algorithm.  Let $G$ be a group of order $n$ with elements $g_1,\cdots,g_n$.  Then I consider each $\langle g_i\rangle $ which are all cyclic subgroups of $G$.  Of course, if $\langle g_i\rangle =\langle g_j\rangle $, we append only $\langle  g_i\rangle $ into our result set.  We call all distinct subgroups of the form $\langle g_i\rangle $ to be the first-generation. (Note that the number of the first-generation subgroups is at most $n$.)  To get $(i+1)$-th generation from $i$-th generation inductively, we construct the subgroup $S$ from each two elements of $i$-th generation $S_1, S_2$ to be  $$S=\langle S_1,S_2\rangle .$$  Obviously, if generated subgroup is already in previous generations, we do not call it $(i+1)$-th generation, hence every subgroups of $G$ has the unique generation number.  We do this construction until the number of the typical generation subgroups to be $0$, since we get all subgroups of $G$.  This works perfectly to any groups, but unfortunately, it is not fast although I implemented additional steps to avoid repeated computation.  So I searched some other algorithms but they are all vague to understand.  One of the algorithm is called the cyclic extension algorithm, but I can not get it easily unless I have studied some backgrounds for this method.  If you have enough knowledge about the cyclic extension algorithm or some other algorithms, please give me an easy explanation. I am in undergrad. course of math dept. Thanks.
34966
I calculated the subgroups of $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ as $\{ (0, 0 ) \}, \{ (0, 0 ), ( 1, 0 ) \}, \{ (0, 0 ), (0, 1) \}, \{ (0, 0 ), (1, 1) \}$ and $ \mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ and itself. I  a bit and found couple of answers which did not include $\{ (0, 0), (1, 1)\}$. Isn't $\{ (0,0), (1,1)\}$ subgroup of  $\mathbb Z / 2 \mathbb Z \times \mathbb Z / 2 \mathbb Z$ ?
36556
This is a homework problem and I'm having trouble just getting a basic understanding.  I understand that $S_6$ is a symmetric group of degree $6$.  I'm not sure how to start looking for elements.
40122
Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.  This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?
70590
How can we prove that "Every Group of order $216$ contains a normal subgroup of order $27$ or $9$ ?".  MY TRY: $216=2^3.3^3$.So possible order of sylow $3$ subgroup is $1$ or $4$ and similarly for sylow $2$ subgroup is of order $1,3,9\,\text{or}\,27$.Then how to proceed $?$Thank you
6113
Suppose we have a set of $n$ elements. We want to know the maximum number of subsets of $3$ elements we can find of this set such that no two subsets have an intersection with $2$ or more elements.  The question arises from trying to count the number of subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$ in a group with $n$ elements of order $2$ such that the product of any two such elements has also order $2$.
93086
Let $G$ be an abelian group of order 72. How many subgroups of order 8 and 4 can it have?    I have listed all possible abelian groups there are 6. Then i said that if I'm looking for an abelian group of order 8 I have 3 possible abelian groups and for each 1 of the 6 first abelian groups must check if 1 of the 3 of the later ones are subgroups. But that is taking forever and I don't have any easy way to check if they are subgroups. I use "If A is a subgroup of G and B is a subgroup of H, then the direct product A × B is a subgroup of G × H". So I found that for every possible G there is 1 subgroup of order 8. But is that correct? If I have a subgroup of a direct product say A × B of a G SxS1xS2x...xSn then A must be subgroup of S and B of S1?
8071
I have tried to solve it for the case when $n=1,2,3,4$ which i found that there is only one group for $n=1$, one group for $n=2,$ one group for$ n=3 $ and two groups for $ n=4$, then my question come, if i were given a number $n$ which is larger, how do i know how many different groups are there?  (I treat two isomorphic groups as the same group in this question.)
73328
For which $n$ is there a group of order $n^2$ without a subgroup of order $n$.  Such groups can not be nilpotent.   This question is related to .
95700
I would like to count subgroups of each order (2, 3, 4, 6, 8, 12) of $S_4$, and, hopefully, convince others that I counted them correctly.  In order to do this by hand in the term exam, I need a clever way to do this because there can be as many subgroups of a group of order 24 as $2^{23}$.  Do you know how to do this?   (I would be most grateful if you could tell me what part of the answer to the old question answers my question before voting to close.)
53023
This is a question in Pinter's A Book of Abstract Algebra.     Let $S=\{g\in G\mid \operatorname{ord}(g)=p\}$. Prove the order of $S$ is a multiple of $p-1$.   In his solution Pinter says $a \in S$ implies that $a$ generates a subgroup with $p-1$ elements. Shouldn't there be $p$ elements $\{1,a^1,\dots,a^{p-1}\}$? Or is it typical to only count the non-trivial elements in a subgroup?
26758
This is a homework question: I need to find three subgroups of the free group with two generators, $F_2$, with certain properties. I have found the other two by constructing covering spaces of $S^1 \vee S^1$ and using fundamental groups; I don't think that this approach works here:     Show that $F_2$ has an infinite index, free subgroup of rank two.   If anyone could nudge me in the right direction I'd be very grateful!
70190
With the next group:  gap&gt; G:=Group((1,2,3,4),(1,2));  Group([ (1,2,3,4), (1,2) ])  gap&gt; Order(G);  24  gap&gt; Elements(G);  [ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4),   (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2),   (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2),   (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ]   Is there any function to list the subgroups of G? Thanks, and sorry if the question looks so easy, i'm a newbie in GAP.
88713
The Question: Determine all groups G such that $ \{ H=(g,g^2): g \in G \}$  is a subgroup of $ G\times G$   I am confused what the question is asking for. I assume there is some small finite number of groups that satisfy this property i.e groups with 4 or less elements.  Things I know:   The subgroup test states that if H is a subgroup then it must contain the identity element of G(say $e$) but since $e^2=e$ then by the criterion of H every cartesian product $G \times G$ contains $(e,e^2)$ Right? If $(g,g^2)$,$(h,h^2)$ $\in$ H then $(gh,(gh)^2)$ $\in$ H. Not sure how to proceed with this. And if $(g,g^2)$ $\in$ H then $(g^{-1},(g^2)^{-1})$ $\in$ H.   Any hints/clarifications would be appreciated.
90245
I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved.
31401
I tried getting $n_{2}$ and $n_{7}$ , denote the number of Sylow-2-Subgroups and Sylow-7-Subgroups respectively.  I got two cases for $n_{2} = 1 , 7 $ and for  $n_{7} = 1 , 8$ , i noticed that if $n_{2} = 1$ and $n_{7}= 1$ ,then we are done since they are unique subgroups and hence will be normal in $G$.  Next how to proceed with the cases of $n_{2} = 7$ and $n_7 = 8$.  Please help?
11828
Let $GL_n(k)$ be the n by n general linear group over $k$, $B_n(k)$ be the subgroup of $GL_n(k)$ consisting of all upper triangular matrices, and $U_n(k)$ be the subgroup of $B_n(k)$ whose diagonal elements are all 1.   To show $B_n(k)$ is solvable, I'm proving it now by following steps: 1. $U_n(k)$ is a subgroup of $B_n(k)$. (done) 2. $U_n(k)$ is normal in $B_n(k)$. (done) 3. $U_n(k)$ is solvable. (question) 4. $B_n(k) / U_n(k)$ is also solvable. (not yet) 5. $B_n(k)$ is solvable. (by the below thm)  I'll use a theorem to verify $B_n(k)$ is solvable.   G is solvable if and only if H and G/H are solvable for some normal subgroup $H$ of $G$.   So, I have to prove both step3 and step4. But I have no idea about them. How to prove? Since my knowledge is not enough, I don't wanna show using Lie theory.  Thanks in advance.