List questions from free-product

Published on 24 Jun 2016 - 15:19

446

Views

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

Published on 10 Jul 2013 - 17:48

404

Views

Decomposability of free product of groups

Published on 31 Jul 2013 - 8:35

185

Views

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

Published on 13 Apr 2017 - 21:45

320

Views

Free product has solvable word problem if its factors do

Published on 24 Apr 2017 - 6:22

128

Views

Certain property of free products of groups

Published on 27 Apr 2017 - 8:56

451

Views

Subgroups of an amalgamated free product of index two

Published on 03 May 2017 - 8:23

250

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Commutator of free products of groups

Published on 04 Jun 2017 - 14:09

369

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

Published on 10 Aug 2017 - 17:12

118

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Clarification on Lemma 68.5 of Munkres

Published on 26 Sep 2017 - 2:37

176

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The relationship between semidirect product and free product

Published on 15 Nov 2017 - 1:26

1.3k

Views

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

Published on 16 Dec 2017 - 18:05

332

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 21 Jun 2011 - 3:19

9.8k

Views

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

Published on 06 Mar 2012 - 20:07

111

Views

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

Published on 07 Sep 2020 - 1:41

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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