Showing that if $|G|=55$, $Hle G$ and $Ntriangleleft G$ with $|H|=5$, $|N|=11$ and $N_G(H)=H$ then $G$ is not...
$begingroup$
I tried to show this by contradiction:
Suppose $Gcong Bbb Z/55Bbb Z cong Bbb Z/11Bbb ZtimesBbb Z/5Bbb Z$ where the second isomorphism results from the fact that $(5,11)=1$
Then $|H|=5$ is prime so if $xinBbb Z/11Bbb Z$ is some element, then $H$ is of the form ${x}timesBbb Z/5Bbb ZcongBbb Z/5Bbb Z $ and $(1+x,1)+Bbb Z/5Bbb Z-(1+x,1)=Bbb Z/5Bbb Z$ but $(1+x)notin H$ so the normalizer strictly contains $H$. In fact all of $G$ is the normalizer, which is a contradiction because $Gne H$
Does this seem ok as a proof?
You might think I didn't use the fact that $N$ is normal. But this is the first point of a 3 point exercise so it might be used later.
group-theory proof-verification finite-groups
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
$endgroup$
add a comment |
$begingroup$
I tried to show this by contradiction:
Suppose $Gcong Bbb Z/55Bbb Z cong Bbb Z/11Bbb ZtimesBbb Z/5Bbb Z$ where the second isomorphism results from the fact that $(5,11)=1$
Then $|H|=5$ is prime so if $xinBbb Z/11Bbb Z$ is some element, then $H$ is of the form ${x}timesBbb Z/5Bbb ZcongBbb Z/5Bbb Z $ and $(1+x,1)+Bbb Z/5Bbb Z-(1+x,1)=Bbb Z/5Bbb Z$ but $(1+x)notin H$ so the normalizer strictly contains $H$. In fact all of $G$ is the normalizer, which is a contradiction because $Gne H$
Does this seem ok as a proof?
You might think I didn't use the fact that $N$ is normal. But this is the first point of a 3 point exercise so it might be used later.
group-theory proof-verification finite-groups
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
$endgroup$
2
$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44
add a comment |
$begingroup$
I tried to show this by contradiction:
Suppose $Gcong Bbb Z/55Bbb Z cong Bbb Z/11Bbb ZtimesBbb Z/5Bbb Z$ where the second isomorphism results from the fact that $(5,11)=1$
Then $|H|=5$ is prime so if $xinBbb Z/11Bbb Z$ is some element, then $H$ is of the form ${x}timesBbb Z/5Bbb ZcongBbb Z/5Bbb Z $ and $(1+x,1)+Bbb Z/5Bbb Z-(1+x,1)=Bbb Z/5Bbb Z$ but $(1+x)notin H$ so the normalizer strictly contains $H$. In fact all of $G$ is the normalizer, which is a contradiction because $Gne H$
Does this seem ok as a proof?
You might think I didn't use the fact that $N$ is normal. But this is the first point of a 3 point exercise so it might be used later.
group-theory proof-verification finite-groups
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
$endgroup$
I tried to show this by contradiction:
Suppose $Gcong Bbb Z/55Bbb Z cong Bbb Z/11Bbb ZtimesBbb Z/5Bbb Z$ where the second isomorphism results from the fact that $(5,11)=1$
Then $|H|=5$ is prime so if $xinBbb Z/11Bbb Z$ is some element, then $H$ is of the form ${x}timesBbb Z/5Bbb ZcongBbb Z/5Bbb Z $ and $(1+x,1)+Bbb Z/5Bbb Z-(1+x,1)=Bbb Z/5Bbb Z$ but $(1+x)notin H$ so the normalizer strictly contains $H$. In fact all of $G$ is the normalizer, which is a contradiction because $Gne H$
Does this seem ok as a proof?
You might think I didn't use the fact that $N$ is normal. But this is the first point of a 3 point exercise so it might be used later.
group-theory proof-verification finite-groups
group-theory proof-verification finite-groups
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
asked Jan 6 at 14:05
John CataldoJohn Cataldo
1,1831316
1,1831316
2
$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44
add a comment |
2
$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44
2
2
$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44
$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44
add a comment |
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$begingroup$
Easier to say: In an abelian group every subgroup is normalised by the whole group.
$endgroup$
– ancientmathematician
Jan 6 at 14:44