$G$ is the dihedral group $D_6$ of order 6. I need to find all subgroups of order 2 and order 3. I have found the cyclic group $\left&lt;b\right&gt; = \{e, b\}$, but that's as far as I have come.  How do you list all subgroups of eg.  $D_6$? and how would you present all subgroups of eg. order 2?
122881
How many subgroups of order $n$ does $D_n$ have?  My work:  Since subgroups of $D_n$ are either cyclic or dihedral, the subgroups of order $n$ of $D_n$ are $\left&lt; r \right&gt;$ (cyclic) and $D_{n/2}$ (dihedral). However, this works only for $n$ is even and $n\geq 2$. I don't know how to do when $n$ is odd.
81440
I am wondering how do we find all subgroups of a direct product? Is there a method to find it?  For example, how can we find all the subgroups of $\mathbb{Z}_2\times\mathbb{Z}_2$?  There is the answer: $\{(0, 0), (1, 0)\}, \{(0, 0), (0, 1)\}, \{(0, 0), (1, 1)\}$, and the improper and trivial subgroup. But I do not know the method how to find it.   All I can think about is the following: since the order of $\mathbb{Z}_2\times\mathbb{Z}_2$ is $4$, the order of a subgroup of it can be $1,2$ or $4$.   If order is $1$, then it is trivial, and if order is $4$, it is proper. Then, the order of a subgroup is $2$. We know all the elements of   $\mathbb{Z}_2\times\mathbb{Z}_2$, so we try all the pairs.   Is that the method?   But what if we are given a bigger order product, such as $\mathbb{Z}_2\times\mathbb{Z}_8\times\mathbb{Z}_3\times\mathbb{Z}_9$? It looks very long to try all possibilities.   Can anyone help?  Thanks
35777
Given the following groups, what is the maximum possible order for an element for   (a) $S_5$ (b) $S_6$ (c) $S_7$ (d) $S_{10}$ (e) $S_{15}$  My book justifies the answer as          (a) The greatest order is $6$ and comes from a product of disjoint cycles of length 2 and 3          (b) The greatest order is $6$ and comes from a cycle of length $6$      The other answers were justified exactly the same way, that is (c) 12, (d) 30, (e) 105  I do not understand how in (a) we even got the number "6" from $S_5$ and what disjoint cycles they are referring to. Could someone at least justify one for me?
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.
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Lemma $1.92$ in Rotman's textbook (Advanced Modern Algebra, second edition) states, Let $G = \langle a \rangle$ be a cyclic group. (i) Every subgroup $S$ of $G$ is cyclic. (ii) If $|G|=n$, then $G$ has a unique subgroup of order $d$ for each divisor $d$ of $n$.  I understand how every subgroup must be cyclic and that there must be a subgroup for each divisor of $d$. But how is that subgroup unique? I'm having trouble understanding this intuitively. For example, if we look at the cyclic subgroup $\Bbb{7}$, we know that there are $6$ elements of order $7$. So we have six different cyclic subgroups of order $7$, right? Thanks in advance.
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Let $G$ be a group. Prove or disprove that $H =\{g^2 | g \in G\}$ is a subgroup of $G$.   I tried testing the permutations of $A_4$, however squaring each cycle yielded a cycle in $A_4$ so I'm lacking a counter-example (if there is one). In a nutshell I'm looking for a subgroup such that when you square the permutation cycle, it yields a cycle not in that subgroup.  Or, I could be way off base and figure out that there isn't a counter-example and I need to prove that indeed $H$ is a subgroup of $G$.
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Is a quotient group a subgroup?   I am trying to solve the question below, and if the answer to above is "yes," then my proof will follow below.   I am trying to prove that a cyclic group G, has a cyclic quotient group G/N.  Where N is a normal subgroup.  A SUBGROUP OF A CYCLIC GROUP IS CYCLIC.  By   So if   (1) G is cyclic   (2) G/N is a subgroup   (3) G/N is a cyclic subgroup   (4) The Quotient group G/N is cyclic   I am unsure about my reasoning in step (2), Is this a valid assumption?
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Let $D_n$ be the dihedral group of order $2n&gt;4$; so it contains a cyclic subgroup $C$ of order $n$ on which $\sigma \in D_n$ outside of $C$ acts as $\sigma c \sigma^{-1}=c^{-1}$ for all $c \in C$. When is the cyclic subgroup $C$ with the above property unique? Determine all $n$ for which $D_n$ has a unique cyclic subgroup $C$ and justify your answer.  Isn't it true that all $D_n$ have a unique cyclic subgroup of order $n$, namely the one that includes all the rotations? And isn't it true that the reflections act the way described above on the rotations? But then the problem seems trivial, so perhaps I'm misunderstanding something?
12153
Show that  the  alternating  group  $A_4$  of all even permutations of $S_4$ does not contain a  subgroup of order $6$.   For me am thinking to write all elements of $A_4$ and trying to find every cyclic subgroup generated by each element of $A_4$, then I have to check whether there exist such a subgroup or not! This is a long procedure for me, I ask if there is a short way to do this.
2919
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?  Why isn't there other possibility?  Thanks :)
36106
Is $D_n$, the dihedral group of order $2n$, isomorphic to a subgroup of $S_n$ ( symmetric group of $n$ letters) for all $n&gt;2$?
55125
Prove that every group of order $35$ is cyclic.    Now, the subgroups of this are ones whose orders divide the order of this group(by lagrange), these are of prime orders $7$ and $5$.  and I guess $\Bbb Z_7\times \Bbb Z_5$ is of order $35$ and since these are both cyclic, so is $\Bbb Z_{35}$.  But that doesn't prove anything about every subgroup of order $35$.  How do I do this.
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I am wondering if my claim is correct, and how to formalize/write it? Thank you.     Give an example of a non-abelian group $G$ containing a proper normal subgroup $N$ such that $G/N$ is abelian.   So I am thinking of the dihedral group of triangle, with $3$ rotations and $2$ reflections. This is not abelian; but if I quotient $3$ rotations, the group has only two elements, which must be abelian.
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Find the three 2-Sylow subgroups of $S_3$ and find a 2-Sylow subgroup and a 3-Sylow subgroup of $S_4.$  I just learned Sylow' theorem at the moment and I don't know how to do these problems. I know a p-Sylow subgroup of a group $G$ is a subgroup of order $p^k$, where $p^k$ divides the order of $G$ but $p^{k+1}$ does not. $|S_3| = 6 = 2\dot\ 3$ and $|S_4| = 24 = 2^3\dot\ 3$. How can I solve these types of problems?
69124
Find all subgroups of $\mathbb{Z}\times\mathbb{Z}$.   I can find all the infinite subgroups of the form $n\mathbb{Z}\times m\mathbb{Z}$, where $n,m$ run over $\mathbb{Z}$. But I don't know how to write out all the finite subgroups in a uniform expression.  Any suggestions? Thanks in advance.
98793
Question is to Prove that $(\mathbb{Z}/2^n \mathbb{Z})^*$ is not  cyclic Group for $n\geq 3$.  Hint : Find two subgroups of order $2$.  I somehow feel that a cyclic group can not have two distinct groups of same order. but, I am not sure about the proof.  I have no idea how to proceed for this.   any hint would be appreciated.  Thank You.
102088
I am trying to understand subgroups.I know a given a group G under a binary operation ∗, a subset H of G is called a subgroup of G if H also forms a group under the operation ∗.   So if I where to list 5 subgroups of $\mathbb{Z}$ could I simply say, $\mathbb{Z}_2$, $\mathbb{Z}_4$, $\mathbb{Z}_6$, $\mathbb{Z}_7$, $\mathbb{Z}_9$, are all subgroups of $\mathbb{Z}$?