List questions from free-product

Published on 30 Apr 2026 - 13:52

449

Views

Showing $\mathbb{Z}_{2} * \mathbb{Z}_{3} \cong\ (a, b\ |\ a^2 = b^3 = e)$

Published on 12 Apr 2026 - 9:02

407

Views

Decomposability of free product of groups

Published on 13 Apr 2026 - 0:54

188

Views

Conjugacy Classes in $\mathbb{Z}*\mathbb{Z}$?

Published on 10 Apr 2026 - 9:54

323

Views

Free product has solvable word problem if its factors do

Published on 13 Apr 2026 - 16:10

131

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Certain property of free products of groups

Published on 15 Apr 2026 - 15:53

454

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Subgroups of an amalgamated free product of index two

Published on 16 Apr 2026 - 1:09

253

Views

Commutator of free products of groups

Published on 15 Apr 2026 - 3:14

372

Views

Is there a non-trivial group $C$ such that $AC \cong BC$ implies $A \cong B$?

Published on 12 Apr 2026 - 12:44

121

Views

Clarification on Lemma 68.5 of Munkres

Published on 16 Apr 2026 - 0:42

179

Views

The relationship between semidirect product and free product

Published on 11 Apr 2026 - 23:46

1.3k

Views

$SL_2(\mathbb{Z})$ is an amalgamated product

Published on 08 Apr 2026 - 16:33

335

Views

If $\phi:G\twoheadrightarrow H$ and if $H=H_1_{H_3}H_2$ for $H_i\leq H$, then is $G=G_1_{G_3}G_2$, for $\phi(G_i)=H_i$?

Published on 15 Apr 2026 - 22:28

9.8k

Views

Difference between free groups, free product of two groups, tensor product of two infinite cyclic groups

Published on 13 Apr 2026 - 18:51

114

Views

Is this the way to embed $G_1$, $G_2$ into the amalgamated product $G_1*_HG_2$?

Published on 17 Apr 2026 - 7:26

List Question

A proposition about free product

Showing $\mathbb{Z}_{2} * \mathbb{Z}_{3} \cong\ (a, b\ |\ a^2 = b^3 = e)$

Decomposability of free product of groups

Conjugacy Classes in $\mathbb{Z}*\mathbb{Z}$?

Free product has solvable word problem if its factors do

Certain property of free products of groups

Subgroups of an amalgamated free product of index two

Commutator of free products of groups

Is there a non-trivial group $C$ such that $AC \cong BC$ implies $A \cong B$?

Clarification on Lemma 68.5 of Munkres

The relationship between semidirect product and free product

$SL_2(\mathbb{Z})$ is an amalgamated product

If $\phi:G\twoheadrightarrow H$ and if $H=H_1_{H_3}H_2$ for $H_i\leq H$, then is $G=G_1_{G_3}G_2$, for $\phi(G_i)=H_i$?

Difference between free groups, free product of two groups, tensor product of two infinite cyclic groups

Is this the way to embed $G_1$, $G_2$ into the amalgamated product $G_1*_HG_2$?

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