How can we find a minimal set of generators of $S_3 rtimes S_3$. [closed]
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I can easily find the generators of direct product two known groups. Is there any way to found the minimal set of generators of the semidirect product of two groups. It can be seen easily that elements like (g,e) and (e,g) generate this group $S_3 rtimes S_3$.
Here the semidirect product is gotten by letting $S_3$ act on itself by conjugation.
abstract-algebra group-theory semidirect-product
closed as off-topic by Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn Nov 16 at 21:36
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I can easily find the generators of direct product two known groups. Is there any way to found the minimal set of generators of the semidirect product of two groups. It can be seen easily that elements like (g,e) and (e,g) generate this group $S_3 rtimes S_3$.
Here the semidirect product is gotten by letting $S_3$ act on itself by conjugation.
abstract-algebra group-theory semidirect-product
closed as off-topic by Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn Nov 16 at 21:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29
add a comment |
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up vote
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I can easily find the generators of direct product two known groups. Is there any way to found the minimal set of generators of the semidirect product of two groups. It can be seen easily that elements like (g,e) and (e,g) generate this group $S_3 rtimes S_3$.
Here the semidirect product is gotten by letting $S_3$ act on itself by conjugation.
abstract-algebra group-theory semidirect-product
I can easily find the generators of direct product two known groups. Is there any way to found the minimal set of generators of the semidirect product of two groups. It can be seen easily that elements like (g,e) and (e,g) generate this group $S_3 rtimes S_3$.
Here the semidirect product is gotten by letting $S_3$ act on itself by conjugation.
abstract-algebra group-theory semidirect-product
abstract-algebra group-theory semidirect-product
edited Nov 20 at 9:57
Jyrki Lahtonen
107k12166364
107k12166364
asked Nov 16 at 6:17
MANI SHANKAR PANDEY
11
11
closed as off-topic by Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn Nov 16 at 21:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn Nov 16 at 21:36
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Travis, José Carlos Santos, amWhy, user10354138, Trevor Gunn
If this question can be reworded to fit the rules in the help center, please edit the question.
Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29
add a comment |
Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29
Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29
add a comment |
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Here I want to know the minimal set of generators of $S_3 rtimes S_3$. Usually, we know that the group $S_3$ can be generated by two elements (12) and (123). Definitely
– MANI SHANKAR PANDEY
Nov 16 at 6:42
Which semi-direct product are you talking about? There are at least three non-isomorphic ones. i) the direct product, ii) the one gotten by letting $S_3$ act on itself by conjugation, and iii) the in-between case where $A_3$ acts trivially, but the 2-cycles all act by one of the order two automorphisms.
– Jyrki Lahtonen
Nov 19 at 6:54
@Jykri Lehtonen Thanks for the reply, here semidirect product is by letting S_3 act on itself by conjugation.
– MANI SHANKAR PANDEY
Nov 20 at 6:29