I'm stuck on the following problem:  Find the smallest number $n$ for which the following statement is true: "The number of isomorphism types of abelian groups with $n$ elements is equal to 10". How many such $n$ are there?  My thoughts: I think that the smallest number is $n=144$. How would I prove this? Also, am I correct in saying that there are infinitely many such numbers $n$?  Any help is appreciated!
62210
Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?   Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.  Can anyone please help  me understand.  Thank you.
31385
I need to show that there are only 2 groups of order 6 up to isomorphism.  I did prove it, but the proof is quite cumbersome. I wonder if there is a very concise proof.  My proof outline: Suppose $G$ is of order 6 and is not $\mathbb{Z}_6$. Then any element that is not identity must have order 2 or 3.  I went on to show that there must be a element of order 2 and another one of order 3, and the intersection of the cyclic subgroups generated by the two is the identity. Then I'm able to show it is isomorphic to $S_3$.
51006
I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.   Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.  Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on?
41051
I have this homework problem.  It states:  By constructing the possible Cayley tables, list all groups where the underlying set has at most 4 elements, up to isomorphisms.  That is, no two groups in your list should be isomorphic.   I understand the basic use of Cayley tables in the sense that they are used to see what the operation on two of the elements becomes.  One group that I know that would fit this would be $\mathbb Z_4$.  I'm not sure how to list all of them and if there is a way to know if it will be isomorphic or not.
78907
This question was asked in my quiz and I am having difficulties in solving it .  Question : For a positive integer $n$ , let $S_n$ denote permutation group on $n$ symbols. Choose correct statements from below. (a) For a positive integer $n$ and for every $m$ with $1\leq m \leq n , S_n $ has a cyclic subgroup of order $m$; (b) For every positive integer $n$ and for every $m$ with $n&lt;m &lt;n!$, $S_n$ has a cyclic subgroup of order $m$; (c) There exists positive integers $n$ and $m$ with $n&lt;m&lt;n!$ such that $S_n$ has a cyclic subgroup of order $m$; (d) For every positive integer $n$ and for every group $G$ of order $n$, $G$ is isomorphic to a subgroup of order $S_n$ .  Attempt: (a) is true as I can choose an element of length $m$ which will generate cyclic subgroup of order $m$. (b) Choose $n=4$ and $m=19$ to contradict (b). (c) I dont know if there can be found any such case and if yes how to do so . (d) I dont know about it . kindly tell. Answer:   A,C,D are true
35999
I wanted to share with you my resolution of this exercise.     How many different groups of order $15$ there are?   My resolution:     We're looking for groups such that $|G|=15=3\cdot 5$. Then: $G$ has an unique Sylow $3$-subgroup of order $3$ ($P_3$) and an unique Sylow $5$-subgroup of order $5$ ($P_5$).      Furthermore, $$|P_3 \cap P_5|\  \Large\mid\ \normalsize|P_3|,\ |P_5|$$   So $$|P_3 \cap P_5|=1\implies P_3 \cap P_5=\{e\}.$$      Then:$$|P_3\cdot P_5|=\displaystyle\frac{|P_3|\cdot |P_5|}{|P_3 \cap P_5|}=|P_3|\cdot |P_5|=3\cdot 5=|G|,$$      and $$G=P_3\cdot P_5 \simeq P_3 \times P_5 \simeq \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}.$$   It's everything correct and well justified? There's some things that I didn't say, for example that a subgroup a prime $p$ as order is cyclic, and thus isomoprhic to $\mathbb{Z}/p\mathbb{Z}$. Maybe I did something wrong. I would appreciate someone to look and criticize my exercise.  Thank you.
18496
In trying to understand why there are two sets of groups of order 4.  I know that there exists the Vierergruppe of $\mathbb{Z}_2 \times \mathbb{Z}_2$ and the group $\mathbb{Z}_4$ but I do not understand why they are not isomorphic
12667
So I'm having trouble with this problem. I know that the definition of a simple group means that the group has no nontrivial subgroups. I know that this can be proven somehow with the help of the converse of Lagrange's Theorem for Abelian groups: If G is abelian of order n, and d is a divisor of n, then G has a subgroup of order d.   My attempt: (=>)Assume that G is a finite abelian group and is simple, then G has no nontrivial normal  subgroups.  (Now I don't know how to show that this implies that G has order p, where p is prime.  (&lt;=)Assume that G is a finite abelian group with order p, where p is a prime. (Since the order of p is prime then what does this mean?)  Edit: Can someone check my new attempt at the proof?   (=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and  is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.   (&lt;=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.
28518
I want to know if there is some systematic way (using some combinatorial argument) to find the number of elements of conjugacy classes of $S_n$ for some given $n$.  For example, let's consider $S_5$. If the representative for the conjugacy class is an $m$-cycle then Dummit and Foote gives a formula on how to compute the number of elements in the conjugacy class. This is not a problem. But what about when the representative is not an $m$-cycle. As an example we can consider the conjugacy class that gives rise by the partition $2+3$ of $5$. A representative for the conjugacy class would be $(1 2)(3 4 5)$. How can I find the number of such elements?.  Question?:      Does $ {5\choose 2}\cdot { 3 \choose 3}\cdot 2$ give me what I want?   Reasoning: For the first parenthesis I need to choose $2$ elements out of $5$ and for the second set of parenthesis I need to choose $3$ out of the remaining $3$ (noting that they can't be repeats). Finally we can permute these two parenthesis in two ways, thus giving me the above number.   Is this reasoning correct?. If not how does one find the number of elements of such conjugacy classes.  As always, any help is greatly appreciated.
42158
What is the smallest positive integer $n$, such that there exist positive integers $a$ and $b$, with $b$ obtained from $a$ by a rearrangement of its digits, so that   $a – b = 11\dots 1$ (The number of '$1$'s equal to $n$)?   Is there any example that can satisfy the equation?
6082
I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one.  Show there is no group homomorphism $f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}$ such that $f(1) = 3$. The first step of the proof is  $|1| = 10$ in $\mathbb{Z}_{10}$, so $|f(1)| \in\{ 1, 2, 5, 10\}$.  Does this mean that $|f(1)|$ can be $1$, $2$, $5$ or $10$? I see that $|1| = 10$ since $1$ is a generator of $\mathbb{Z}_{10}$, and that $|f(1)|$ seems to be a factor of $10$. If $f(1)$ is the image of $1$ under $f$, how would I find the order of $f(1)$?  Thanks for your help.
82188
I recently took a course on group theory, which mentioned that the following proposition is equivalent to the continuum hypothesis: "The infinite symmetric group (i.e. the group of permutations on the set $\mathbb{N}$) has exactly 4 normal subgroups." Does anyone have any references or explanation for this?
87542
Let $G$ be  a group of order  $21$. Find the number of non-isomorphic groups of order $21$  My solution: If  the group $G$ is commutative,then $G$ can be expressed as a direct product of cyclic groups of prime power order,i.e it is either $\mathbb  Z_3\times \mathbb Z_7$ or $\mathbb Z_{21}$ which is same  My problem is when $G$ is commutative.How to proceed here?
49891
I want to prove at any group of order $4$ is isomorphic to either $\Bbb Z_4$ or $\Bbb Z^*_8$. I know that these two groups are not isomorphic to each other because they have different order, but I cannot go further to prove at any group of order 4 should be isomorphic to either of them. Help would be greatly appreciated.  Thanks!
42066
My question is: Can you find a cyclic group with n generators?  I know that zero (or any other identity element for that matter) is included, so there would be for $Z_n$ at most n-1 generators. However, is it possible to say that $Z_{n+1}$ could provide n generators?  Any help would be much appreciated! If you wouldn't mind, I would appreciate it if you provide enough detail to really help me understand so I can learn. Thanks in advance!
21732
How many isomorphisms are there from $\Bbb Z_{12}$ to $\Bbb Z_{4} \oplus \Bbb Z_{3}$?  I can see that if $f$ is the isomorphism, the isomorphism is determined by the value of $f(1)$ because $f(1) = (1,1)$ and $f(k) = f(k) + f(1) = f(k) + (1,1)$.  But since $\Bbb Z_{12}$ has $4$ generators, there are $4$ different ways to set $f(1)$ equal to a generator.  Therefore there are $4$ isomorphisms.  Is this correct?
19132
I will quote a question from my textbook, to prevent misinterpretation:     Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.   I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian.