By Remak's theorem, this is true for finite groups, and so any counter example would counter Remak so is probably nontrivial.
91143
In , The Chaz asks whether $G\times G\cong H\times H$ implies that $G\cong H$, where $G$ and $H$ are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta.  Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism $G\to H$ somehow appealing to the universal property of $\oplus$, and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required.  This question may be considered trivial or weird for someone who's fluent with categories, I don't know.  contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight.  -- edit, So a more precise title would have been "Under what conditions that can be expressed in a universal way does $G\oplus G\cong H\oplus H$ imply $G\cong H$ ?" but I don't like long titles.  -- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously.
80493
If the pair $(X,A)$ is good, then of course we have that $H_n(X,A) \approx \tilde{H}_n(X/A)$. For a counterexample to the converse, what is a nice example of a reasonably non-pathological bad pair $(X,A)$ such that $H_n(X,A) \not \approx  \tilde{H}_n(X/A)$ ?   I would appreciate if $n = 1,2$ (so it can be visualized).
24728
I know it is possible for a group $G$ to have normal subgroups $H, K$, such that $H\cong K$ but $G/H\not\cong G/K$, but I couldn't think of any examples with $G$ finite. What is an illustrative example?
77790
Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.   is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$  If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.  The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$.
53988
I'm trying to prove the third Isomorphism theorem as stated below  Theorem. Let $G$ be a group, $K$ and $N$ are normal subgroups of $G$ with $K⊆N$. Then $$(G⁄K)⁄(N⁄K)≅G⁄N.$$  I look up for some answered on google, but I don't understand any of those. I wonder if any one can show me this proof step by step. I don't want to get it done, I want to understand why this theorem is true.
67383
Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism?   If so why does the 2nd example in  says that the unique non trivial automorphism is negation?
69131
I'm considering finite groups only. Cayley's theorem says the a group $G$ is isomorphic to a subgroup of $S_{|G|}$. I think it's interesting to ask for smaller values of $n$ for which $G$ is a subgroup of $S_n$. Obviously, it's not always possible to do better than Cayley's theorem. But sometimes it is possible (for example, $\mathbb{Z}_6$ as a subgroup of $S_5$).  So I'm asking:        Given a finite group $G$, is there an algorithmic way to find or approximate the minimal $n$ for which $G$ is isomorphic to a subgroup of $S_n$?   If the answer to $(1)$ is not known, is it known for specific classes of groups?   In particular, for finite abelian groups, is it true that for a prime $p$, the minimal $n$ for $\mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}}$ is $p^{t_1}+p^{t_2}$ (I can prove that is is true for different primes $p_1$ and $p_2$, but have problems when it's the same prime in both factors).      Thanks!
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.
122686
Given an integer $n\geq 1$, it is known that there are only finitely many finite groups with exactly $n$ conjugacy classes.  Question: For $n\geq 3$, does there exists a finite non-abelian group, with exactly $n$ conjugacy classes?
40359
It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).  Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?  If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?  And if not, which groups arise as counterexamples?  And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?
8831
There is a famous Theorem telling that:  For $n≥5$, $A_n$ is the only proper nontrivial normal subgroup of $S_n$.  For the proof, we firstly start with assuming a subgroup of $S_n$ which $1≠N⊲S_n$. We proceed until at the last part of proof's body, we assume $N∩A_n=\{1\}$. This assumption should be meet a contradiction with normality of $N$ in $S_n$. There; we get $N=\{1,\pi $} in which $\pi$ is an odd permutation of order $2$. Now for meeting desire inconsistency, I have two approaches:  (a) Since every normal subgroup, having two elements, lies in the center of $G$ so, our $N⊆ Z(S_n)=\{1\}$ for $n≥5$ and then $N=\{1\}$.  (b) Clearly, $1≠N$ acting on set $\Omega=\{1,2,...,n\}$ is intransitive wherein $|\Omega|≥5$ and according to the following Proposition $S_n$ would be imprimitive.  Proposition 7.1: If the transitive group $G$ contains an intransitive normal subgroup different from $1$, then $G$ is imprimitive (Finite Permutation Groups by H.Wielandt).  May I ask if the second approach is valid? I am fond of knowing new approach if exists. Thanks.
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
48606
What is an example of a group $G$ with an element $g$ such that no automorphism of $G$ takes $g$ to its inverse?
60891
If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?  Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.  I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?
20566
Let $X$ be a transitive $G$-set. ($G$ acts on  $X$ transitively.) If $X$ is finite and has at least two elements, show that there is some element $g$ $\in$ G which does not have any fixed points; that is, such that $g$$($$x$$)$ $\ne$ $x$ for all $x$ $\in$ $X$  I am trying to use contradiction, but it is not very clear to me why this is true.
112283
If it is not true, can you provide a counter-example?
23247
Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.  I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true?