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Questions about the branch of algebra that deals with groups.

7 votes
Accepted

Zero divisors of the form $1+x+y$ in the rational group algebra

If $G$ has a cyclic quotient $G/N$ of order $3$ in which $x$ and $y$ map to the two non-identity elements, then $1+x+y$ acts as zero on the non-trivial linear complex representations of $G/N$. So $1+x …
Jeremy Rickard's user avatar
6 votes
Accepted

Cardinality of factors of infinite non-abelian groups

Let $G=C_2\ast C_3$ be the free product of cyclic groups $C_2=\langle a\rangle$ and $C_3=\langle b\rangle$. Let $A$ be the subset of $G$ consisting of $$b,bab,babab,\dots$$ together with all reduced …
Jeremy Rickard's user avatar
3 votes
Accepted

Does every nearly normal subgroup contain a normal subgroup?

Let $K$ be an infinite product of cyclic groups of order 2, and $H$ an index two subgroup of $K$. Let $A$ be the full automorphism group of $K$, and $G=K\rtimes A$. Then since $G$ acts transitively o …
Jeremy Rickard's user avatar
4 votes

Permutation characters and regular orbits

Not every non-cyclic finite group has such a relation. E.g., the quaternion group of order 8 doesn't. (I meant this as a comment on Alex's answer, but I don't have enough reputation to comment.)
Jeremy Rickard's user avatar
6 votes
Accepted

The number in the join of conjugate class and centralizer

Take $G$ to be a symmetric group of degree at least 5, and $a$ a transposition.
Jeremy Rickard's user avatar
2 votes

Classes of groups with finitely many retracts

Any simple group $G$ only has $G$ and the trivial subgroup as retracts.
Jeremy Rickard's user avatar
6 votes

What is "van Dyck's theorem"

I found several statements quickly by Googling (although they varied a bit on the van/von question). The exact formulation varied, but basically it's just the statement that if $G$ is a group given …
Jeremy Rickard's user avatar
8 votes

Groups like symmetric group

How about $D_{n(n-1)}\times C_{n-2}\times S_{n-3}$ for any $n$ where 3 divides $n(n-1)$ but 4 doesn't?
Jeremy Rickard's user avatar
5 votes
Accepted

Modules "projective in a subcategory"

More generally, the following is true: Let $A$ be a finite-dimensional $k$-algebra ($k$ a field), $M$ a finite-dimensional indecomposable (right) $A$-module, and $S$ the category of coproducts of cop …
Jeremy Rickard's user avatar
35 votes
Accepted

Does the symmetric group on an infinite set have a minimal generating set?

I think it follows from Theorem 1.1 of "Subgroups of Infinite Symmetric Groups" by Macpherson and Neumann (J. London Math. Soc. (1990) s2-42 (1): 64-84) that there is no minimal generating set of $S(\ …
Jeremy Rickard's user avatar
17 votes
Accepted

A group whose automorphism group is cyclic

There's a construction of a rank two (and therefore not locally cyclic) abelian group with endomorphism ring $\mathbb{Z}$, and therefore automorphism group cyclic of order 2, in "On the cancellation …
Jeremy Rickard's user avatar
6 votes

New class of finite groups?

Let $G=(C_7\rtimes C_3)\times(C_5\rtimes C_2)$. Then $G$ is supersolvable but doesn't have a normal subgroup of order $7\times 2$ or $3\times 5$.
Jeremy Rickard's user avatar
9 votes

$\left[x,y\right]$ as a product of palindromes of even length?

Let $g=(1,2)$, $h=(2,3)$ two non-commuting involutions in (say) the symmetric group $S_3$. Consider the homomorphism $\varphi:F_2\to S_3$ with $\varphi(x)=g$, $\varphi(y)=h$. If $w$ is any palindromi …
Jeremy Rickard's user avatar
7 votes
Accepted

Given a filtration of a finitely generated module over a noetherian ring that "looks" split,...

Take the direct sum of the short exact sequences $$0\to F_{i+1}\to F_i\to F_i/F_{i+1}\to0$$ for $0\leq i\leq n$. This has the form $$0\to \bigoplus_{i=1}^n F_i\to \bigoplus_{i=0}^n F_i\to F_0\to 0$$ …
Jeremy Rickard's user avatar
10 votes
Accepted

Number of subgroups of a $p$-group of index $p^k$

It seems that the $p>2$ part of this was proved in Kulakoff, A., Über die Anzahl der eigentlichen Untergruppen und der Elemente von gegebener Ordnung in $p$-Gruppen., Math. Ann. 104, 778-793 (1931). Z …
Jeremy Rickard's user avatar

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