Linked Questions
15 questions linked to/from When is $A$ isomorphic to $A^3$?
6 votes
2 answers
2k views
If $G \times G \cong H \times H$, then is $G \cong H$? [duplicate]
Let $G,H$ groups. Suppose that $G \times G \cong H \times H$. Then is necessarily $G \cong H$? I know that if $G,H$ are finite groups then it is true (you show that the map $FinGrp \rightarrow \...
0 votes
1 answer
255 views
Groups with $G^n \cong G$ for some integer $n$ [duplicate]
Which integers $n>2$ have the following property? There is a group $G$ such that $G^n \cong G$; and for all integers $k$ with $1<k<n$ we have $G^k\not \cong G$.
2 votes
0 answers
56 views
About direct products of groups [duplicate]
Let $G$ be a group. Suppose that $G\simeq G\times G\times G$ (here $\simeq$ is an isomorphism of groups). Is it true that in this case $G\simeq G\times G$? Of course, this question is slightly ...
77 votes
28 answers
10k views
Results from abstract algebra which look wrong (but are true)
There are many statements in abstract algebra, often asked by beginners, which are just too good to be true. For example, if $N$ is a normal subgroup of a group $G$, is $G/N$ isomorphic to a subgroup ...
96 votes
2 answers
8k views
$A$ is isomorphic to $A \oplus \mathbb{Z}^2$, but not to $A \oplus \mathbb{Z}$
Are there abelian groups $A$ with $A \cong A \oplus \mathbb{Z}^2$ and $A \not\cong A \oplus \mathbb{Z}$?
18 votes
2 answers
1k views
An order type $\tau$ equal to its power $\tau^n, n>2$
(This is a re-post of my old unanswered question from Math.SE) For purposes of this question, let's concern ourselves only with linear (but not necessarily well-founded) order types. Recall that: $0,...
7 votes
2 answers
688 views
Non-isomorphic groups with isomorphic nth powers (and similarly in other categories)
What's the simplest example (if any) of two non-isomorphic groups G and H such that $G \times G \cong H \times H$? A similar question can be asked for $n^{th}$ powers for fixed $n > 1$. The Krull-...
8 votes
2 answers
457 views
Space $X$ such that $X^\lambda\cong X$ for some $\lambda$
Which cardinals $\lambda > 2$ have the following property? There is a space $(X,\tau)$ such that for all cardinals $\kappa$ with $1<\kappa<\lambda$ we have $X\not\cong X^\kappa$, and $X\cong ...
7 votes
1 answer
2k views
Rank versus free-rank of a module
Suppose $M$ is a finitely generated left module over a ring $R.$ We define the rank of $M$ as the minimal number of generators of $M.$ If in addition $M$ is free, then we define the free-rank of $...
0 votes
1 answer
432 views
Examples of groups such that order isomorphism of the subgroups of $G\times G$ and $H\times H$ does not imply isomorphism of $G$ and $H$
Let $G$ and $H$ be groups, $\operatorname{Sub}(G\times G)$ be the set of all subgroups of $G\times G$ and $\operatorname{Sub}(H\times H)$ be the set of all subgroups of $H\times H$. Assume there ...
1 vote
1 answer
199 views
$n$-product-periodic topological spaces
We call an topological space $(X,\tau)$ $n$-product-periodic for an integer $n\geq 3$ if $\prod_{i=1}^n X \cong X$ but for all integers $k$ with $2\leq k\leq n-1$ we have $\prod_{i=1}^k X \not\cong X$....
6 votes
0 answers
542 views
Square and cube?
Gowers and Maurey proved in their remarkable paper(s), that there is a Banach space $X$ such that $X$ is isomorphic to its cube $X\oplus X\oplus X$ but not to isomorphic to its square $X\oplus X$. ...
4 votes
0 answers
340 views
A Group $G$ with $G \cong G \times G \times G$ and $G \not \cong G \times G$? [duplicate]
Possible Duplicate: when is A isomorphic to A^3? Does there exist a group $G$ such that $G \cong G \times G \times G$ and $G \not \cong G \times G$? If such groups exist, can $G$ be countable? ...
8 votes
0 answers
338 views
Are there two non-isomorphic finitely presented groups which are retracts of each other?
According to answers to this Math Overflow question, there is an infinite rank abelian group $A$ such that $A\cong A^3$ but $A\not\cong A^2.$ Clearly $A$ is an retract of $A^2$ while $A^2$ is an ...
5 votes
0 answers
408 views
A question about retracts of a group
A group $H$ is called a retract of a group $G$ if there exist homomorphisms $f:H\longrightarrow G$ and $g:G\longrightarrow H$ such that $g\circ f=id_H$. By a trivial retract of $G$, I just mean the ...