Newbetuts

How many groups of order at most $25$ are "pleasant" (abelian, with every non-identity element having prime order)?

How to find the order of a Rubik's cube algorithm?

Normal subgroup: a problem in Verification of equivalence

Identity element in a finite cyclic group $G$.

A commutative group structure on $R\times R$ for a ring $R$

Galois group of a degree 5 irreducible polynomial with two complex roots.

Modular subgroup lattice in GAP

How do you compute group cohomology in practice?

Normalizer and centralizer are equivalent when $p$ is the smallest prime dividing $|G|$

If N and every subgroup of N is normal in G then G/N is abelian .

Semidirect product: general automorphism always results in a conjugation

Can a free group over a set be constructed this way (without equivalence classes of words)?

Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f$.

If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$.

$\mathrm{Aut}(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)$ for $G$ a non-abelian simple group

New posts in group-theory

Find $\mathrm{Aut}(G)$, $\mathrm{Inn}(G)$ and $\mathrm{Aut}(G)/\mathrm{Inn}(G)$ for $G = D_4$

Find a combinatorial proof to 10!=7!6!

closed subgroup on a topological group

When does there exist a section of $GL_n(\mathbb Z_p) \rightarrow GL_n(\mathbb F_p)$?

Is a HNN extension of a virtually torsion-free group virtually torsion-free?

How many groups of order at most $25$ are "pleasant" (abelian, with every non-identity element having prime order)?

How to find the order of a Rubik's cube algorithm?

Normal subgroup: a problem in Verification of equivalence

Identity element in a finite cyclic group $G$.

A commutative group structure on $R\times R$ for a ring $R$

Galois group of a degree 5 irreducible polynomial with two complex roots.

Modular subgroup lattice in GAP

How do you compute group cohomology in practice?

Normalizer and centralizer are equivalent when $p$ is the smallest prime dividing $|G|$

If N and every subgroup of N is normal in G then G/N is abelian .

Semidirect product: general automorphism always results in a conjugation

Can a free group over a set be constructed this way (without equivalence classes of words)?

Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f$.

If $|G|=p^3q^2$ then $\Phi(G)$ is cyclic for primes $p\neq q$.

$\mathrm{Aut}(\mathrm{Aut}(G))\cong\mathrm{Aut}(G)$ for $G$ a non-abelian simple group