On groups with $operatorname{Inn}(G) cong D_8$ or $A_4$. [closed]
Let $G/Z(G) cong D_8$, where $D_8$ is the dihedral group of order $8$ and $Z(G)$ is the center of $G$. What are the possibilities for the group $G$ and does there always exist a non-inner automorphism group $G$?
Same question with $A_4$ in place of $D_8$.
Thanks.
group-theory finite-groups
edited Nov 26 at 11:51
the_fox
2,41411431
asked Nov 26 at 10:47
Ashish
141
closed as off-topic by Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo Nov 29 at 23:17
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Let $G/Z(G) cong D_8$, where $D_8$ is the dihedral group of order $8$ and $Z(G)$ is the center of $G$. What are the possibilities for the group $G$ and does there always exist a non-inner automorphism group $G$?
Same question with $A_4$ in place of $D_8$.
Thanks.
group-theory finite-groups
edited Nov 26 at 11:51
the_fox
2,41411431
asked Nov 26 at 10:47
Ashish
141
closed as off-topic by Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo Nov 29 at 23:17
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." – Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $G/Z(G) cong D_8$, where $D_8$ is the dihedral group of order $8$ and $Z(G)$ is the center of $G$. What are the possibilities for the group $G$ and does there always exist a non-inner automorphism group $G$?
Same question with $A_4$ in place of $D_8$.
Thanks.
group-theory finite-groups
edited Nov 26 at 11:51
the_fox
2,41411431
asked Nov 26 at 10:47
Ashish
141
Let $G/Z(G) cong D_8$, where $D_8$ is the dihedral group of order $8$ and $Z(G)$ is the center of $G$. What are the possibilities for the group $G$ and does there always exist a non-inner automorphism group $G$?
Same question with $A_4$ in place of $D_8$.
Thanks.
group-theory finite-groups
group-theory finite-groups
edited Nov 26 at 11:51
the_fox
2,41411431
asked Nov 26 at 10:47
Ashish
141
edited Nov 26 at 11:51
the_fox
2,41411431
asked Nov 26 at 10:47
Ashish
141
edited Nov 26 at 11:51
the_fox
2,41411431
edited Nov 26 at 11:51
the_fox
2,41411431
edited Nov 26 at 11:51
the_fox
2,41411431
2,41411431
asked Nov 26 at 10:47
Ashish
141
asked Nov 26 at 10:47
Ashish
141
asked Nov 26 at 10:47
Ashish
141
141
closed as off-topic by Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo Nov 29 at 23:17
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." – Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo Nov 29 at 23:17
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." – Saad, Paul Plummer, amWhy, user10354138, Davide Giraudo
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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There are infinitely many such groups. For example, $G / Z(G) cong D_8$ when $G cong A times D_{16}$ with $A$ any abelian group.
Similarly, $G / Z(G) cong A_4$ when $G cong B times operatorname{SL}_2(3)$ with $B$ any abelian group.
The "right question" to ask here is for those groups $G$ such that $G / Z(G) cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.
answered Nov 26 at 11:13
the_fox
2,41411431
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For the the dihedral group of order 8 with presentation $<x, y| x^2=y^2=(xy)^4=1>$
consider the group $G=<x, y| x^2=y^2=(xy)^4>$. $G$ is of order 48 with center of order 6, generated by $x^2$.
For the group A_4 with presentation $<x, y| x^2=y^3=(xy)^3=1>$ consider the group
$G=<x, y| x^2=y^3=(xy)^3>$ of order 72 with center of order 6, generated by $x^2$.
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are infinitely many such groups. For example, $G / Z(G) cong D_8$ when $G cong A times D_{16}$ with $A$ any abelian group.
Similarly, $G / Z(G) cong A_4$ when $G cong B times operatorname{SL}_2(3)$ with $B$ any abelian group.
The "right question" to ask here is for those groups $G$ such that $G / Z(G) cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.
answered Nov 26 at 11:13
the_fox
2,41411431
add a comment |
There are infinitely many such groups. For example, $G / Z(G) cong D_8$ when $G cong A times D_{16}$ with $A$ any abelian group.
Similarly, $G / Z(G) cong A_4$ when $G cong B times operatorname{SL}_2(3)$ with $B$ any abelian group.
The "right question" to ask here is for those groups $G$ such that $G / Z(G) cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.
answered Nov 26 at 11:13
the_fox
2,41411431
add a comment |
There are infinitely many such groups. For example, $G / Z(G) cong D_8$ when $G cong A times D_{16}$ with $A$ any abelian group.
Similarly, $G / Z(G) cong A_4$ when $G cong B times operatorname{SL}_2(3)$ with $B$ any abelian group.
The "right question" to ask here is for those groups $G$ such that $G / Z(G) cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.
answered Nov 26 at 11:13
the_fox
2,41411431
There are infinitely many such groups. For example, $G / Z(G) cong D_8$ when $G cong A times D_{16}$ with $A$ any abelian group.
Similarly, $G / Z(G) cong A_4$ when $G cong B times operatorname{SL}_2(3)$ with $B$ any abelian group.
The "right question" to ask here is for those groups $G$ such that $G / Z(G) cong D_8$ (or $A_4$), where $G$ has no abelian direct factors. I do not know the answer to that question. You should have a look at capable groups.
answered Nov 26 at 11:13
the_fox
2,41411431
edited Nov 26 at 12:01
answered Nov 26 at 11:13
the_fox
2,41411431
answered Nov 26 at 11:13
the_fox
2,41411431
answered Nov 26 at 11:13
the_fox
2,41411431
2,41411431
add a comment |
add a comment |
For the the dihedral group of order 8 with presentation $<x, y| x^2=y^2=(xy)^4=1>$
consider the group $G=<x, y| x^2=y^2=(xy)^4>$. $G$ is of order 48 with center of order 6, generated by $x^2$.
For the group A_4 with presentation $<x, y| x^2=y^3=(xy)^3=1>$ consider the group
$G=<x, y| x^2=y^3=(xy)^3>$ of order 72 with center of order 6, generated by $x^2$.
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
add a comment |
For the the dihedral group of order 8 with presentation $<x, y| x^2=y^2=(xy)^4=1>$
consider the group $G=<x, y| x^2=y^2=(xy)^4>$. $G$ is of order 48 with center of order 6, generated by $x^2$.
For the group A_4 with presentation $<x, y| x^2=y^3=(xy)^3=1>$ consider the group
$G=<x, y| x^2=y^3=(xy)^3>$ of order 72 with center of order 6, generated by $x^2$.
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
add a comment |
For the the dihedral group of order 8 with presentation $<x, y| x^2=y^2=(xy)^4=1>$
consider the group $G=<x, y| x^2=y^2=(xy)^4>$. $G$ is of order 48 with center of order 6, generated by $x^2$.
For the group A_4 with presentation $<x, y| x^2=y^3=(xy)^3=1>$ consider the group
$G=<x, y| x^2=y^3=(xy)^3>$ of order 72 with center of order 6, generated by $x^2$.
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
For the the dihedral group of order 8 with presentation $<x, y| x^2=y^2=(xy)^4=1>$
consider the group $G=<x, y| x^2=y^2=(xy)^4>$. $G$ is of order 48 with center of order 6, generated by $x^2$.
For the group A_4 with presentation $<x, y| x^2=y^3=(xy)^3=1>$ consider the group
$G=<x, y| x^2=y^3=(xy)^3>$ of order 72 with center of order 6, generated by $x^2$.
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
answered Nov 26 at 14:26
i. m. soloveichik
3,70511125
3,70511125
add a comment |
add a comment |
