on the order of derived subgroup
Clash Royale CLAN TAG#URR8PPP
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Let $P=M_p^n+1=langle a,b mid a^p^n=b^p=1, a^b=a^1+p^n-1rangle$.
I want to prove that $P^prime=langle [a,b] rangle$.
My Try: Clearly $langle [a,b] rangle le P^prime$. Put $K:=langle [a,b ]rangle$. Since $a^b=a^1+p^n-1$, we have
$[a,b]=a^p^n-1$. Thus $K rm char langle a rangle trianglelefteq P$, and so $K trianglelefteq P$. Consider $P/K$. If $P/K$ is abelian, then we do. Please help me to complete it.
group-theory combinatorial-group-theory
edited Aug 26 at 11:47
Shaun
7,43692972
asked Aug 26 at 10:27
Little girl
1768
add a comment |Â
up vote
3
down vote
favorite
Let $P=M_p^n+1=langle a,b mid a^p^n=b^p=1, a^b=a^1+p^n-1rangle$.
I want to prove that $P^prime=langle [a,b] rangle$.
My Try: Clearly $langle [a,b] rangle le P^prime$. Put $K:=langle [a,b ]rangle$. Since $a^b=a^1+p^n-1$, we have
$[a,b]=a^p^n-1$. Thus $K rm char langle a rangle trianglelefteq P$, and so $K trianglelefteq P$. Consider $P/K$. If $P/K$ is abelian, then we do. Please help me to complete it.
group-theory combinatorial-group-theory
edited Aug 26 at 11:47
Shaun
7,43692972
asked Aug 26 at 10:27
Little girl
1768
3
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $P=M_p^n+1=langle a,b mid a^p^n=b^p=1, a^b=a^1+p^n-1rangle$.
I want to prove that $P^prime=langle [a,b] rangle$.
My Try: Clearly $langle [a,b] rangle le P^prime$. Put $K:=langle [a,b ]rangle$. Since $a^b=a^1+p^n-1$, we have
$[a,b]=a^p^n-1$. Thus $K rm char langle a rangle trianglelefteq P$, and so $K trianglelefteq P$. Consider $P/K$. If $P/K$ is abelian, then we do. Please help me to complete it.
group-theory combinatorial-group-theory
edited Aug 26 at 11:47
Shaun
7,43692972
asked Aug 26 at 10:27
Little girl
1768
Let $P=M_p^n+1=langle a,b mid a^p^n=b^p=1, a^b=a^1+p^n-1rangle$.
I want to prove that $P^prime=langle [a,b] rangle$.
My Try: Clearly $langle [a,b] rangle le P^prime$. Put $K:=langle [a,b ]rangle$. Since $a^b=a^1+p^n-1$, we have
$[a,b]=a^p^n-1$. Thus $K rm char langle a rangle trianglelefteq P$, and so $K trianglelefteq P$. Consider $P/K$. If $P/K$ is abelian, then we do. Please help me to complete it.
group-theory combinatorial-group-theory
edited Aug 26 at 11:47
Shaun
7,43692972
asked Aug 26 at 10:27
Little girl
1768
edited Aug 26 at 11:47
Shaun
7,43692972
edited Aug 26 at 11:47
Shaun
7,43692972
edited Aug 26 at 11:47
Shaun
7,43692972
7,43692972
asked Aug 26 at 10:27
Little girl
1768
asked Aug 26 at 10:27
Little girl
1768
asked Aug 26 at 10:27
Little girl
1768
1768
3
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34
add a comment |Â
3
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34
3
3
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34
add a comment |Â
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3
The images of the generators $a$ and $b$ of $P$ commute in $P/K$, because $[a,b] in K$. If the generators of a group commute with each other, then the group is abelian.
– Derek Holt
Aug 26 at 10:32
@Derek Holt Thanks. Really, $P/K=langle aK,bK rangle$. Thats right?
– Little girl
Aug 26 at 10:34