Proof that a group of order 8888 is solvable [closed]
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I need to prove that every group of order $8888$ is solvable by proving that $G^{(3)}={e}$. Can you give some help?
group-theory finite-groups solvable-groups
edited Dec 21 '18 at 16:55
Shaun
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asked Dec 19 '18 at 12:01
orisofer2orisofer2
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closed as off-topic by Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste Dec 19 '18 at 16:32
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$begingroup$
I need to prove that every group of order $8888$ is solvable by proving that $G^{(3)}={e}$. Can you give some help?
group-theory finite-groups solvable-groups
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
$endgroup$
closed as off-topic by Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste Dec 19 '18 at 16:32
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I need to prove that every group of order $8888$ is solvable by proving that $G^{(3)}={e}$. Can you give some help?
group-theory finite-groups solvable-groups
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
$endgroup$
I need to prove that every group of order $8888$ is solvable by proving that $G^{(3)}={e}$. Can you give some help?
group-theory finite-groups solvable-groups
group-theory finite-groups solvable-groups
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
edited Dec 21 '18 at 16:55
Shaun
10.1k113685
10.1k113685
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
asked Dec 19 '18 at 12:01
orisofer2orisofer2
22
22
closed as off-topic by Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste Dec 19 '18 at 16:32
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste Dec 19 '18 at 16:32
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Adrian Keister, Paul Frost, Dietrich Burde, Namaste
If this question can be reworded to fit the rules in the help center, please edit the question.
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1 Answer
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Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 cdot 11 cdot 101$. If $P in Syl_{101}(G)$, prove that $P lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $gt 1$ divisors of $8$). If $Q in Syl_{11}(G)$, this amounts to $PQ lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 cdot 11 cdot 101$. If $P in Syl_{101}(G)$, prove that $P lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $gt 1$ divisors of $8$). If $Q in Syl_{11}(G)$, this amounts to $PQ lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
$endgroup$
add a comment |
$begingroup$
Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 cdot 11 cdot 101$. If $P in Syl_{101}(G)$, prove that $P lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $gt 1$ divisors of $8$). If $Q in Syl_{11}(G)$, this amounts to $PQ lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
$endgroup$
add a comment |
$begingroup$
Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 cdot 11 cdot 101$. If $P in Syl_{101}(G)$, prove that $P lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $gt 1$ divisors of $8$). If $Q in Syl_{11}(G)$, this amounts to $PQ lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
$endgroup$
Maybe you should display some efforts of your own. Let me nevertheless give some hints / one way to do this. $8888=2^3 cdot 11 cdot 101$. If $P in Syl_{101}(G)$, prove that $P lhd G$, since all divisors of $88$ greater than $1$ are smaller than $101$. Look at $G/P$, and show that its Sylow $11$-subgroup is normal (also by inspecting the $gt 1$ divisors of $8$). If $Q in Syl_{11}(G)$, this amounts to $PQ lhd G$. Now use that $PQ$ is cyclic of order $1111$ (and index $8$) and that any group of order $8$ has a trivial second derived subgroup.
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
edited Dec 19 '18 at 13:28
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
answered Dec 19 '18 at 13:21
Nicky HeksterNicky Hekster
29.1k63456
29.1k63456
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