Proving the chain A/B≅A/C≅X
$begingroup$
Consider the following groups:
$$ A=mathbb{Z}_{4}timesmathbb{Z}_{2},
B={0,2}timesmathbb{Z}_{2},
C=mathbb{Z}_{4}times{0} $$
I would like to prove:
$A/Bcong A/C$.
$Bnot cong C$.
In order to do so, I need to find $X$ so $A/Bcong A/Ccong X$. At start I though that $X=mathbb{Z}_2$ but I can't to seem to prove that $A/Bcong X$. How to find $X$?
abstract-algebra group-theory
asked Jan 12 at 18:14
abuka123abuka123
273
$endgroup$
add a comment |
$begingroup$
Consider the following groups:
$$ A=mathbb{Z}_{4}timesmathbb{Z}_{2},
B={0,2}timesmathbb{Z}_{2},
C=mathbb{Z}_{4}times{0} $$
I would like to prove:
$A/Bcong A/C$.
$Bnot cong C$.
In order to do so, I need to find $X$ so $A/Bcong A/Ccong X$. At start I though that $X=mathbb{Z}_2$ but I can't to seem to prove that $A/Bcong X$. How to find $X$?
abstract-algebra group-theory
asked Jan 12 at 18:14
abuka123abuka123
273
$endgroup$
$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15
add a comment |
$begingroup$
Consider the following groups:
$$ A=mathbb{Z}_{4}timesmathbb{Z}_{2},
B={0,2}timesmathbb{Z}_{2},
C=mathbb{Z}_{4}times{0} $$
I would like to prove:
$A/Bcong A/C$.
$Bnot cong C$.
In order to do so, I need to find $X$ so $A/Bcong A/Ccong X$. At start I though that $X=mathbb{Z}_2$ but I can't to seem to prove that $A/Bcong X$. How to find $X$?
abstract-algebra group-theory
asked Jan 12 at 18:14
abuka123abuka123
273
$endgroup$
Consider the following groups:
$$ A=mathbb{Z}_{4}timesmathbb{Z}_{2},
B={0,2}timesmathbb{Z}_{2},
C=mathbb{Z}_{4}times{0} $$
I would like to prove:
$A/Bcong A/C$.
$Bnot cong C$.
In order to do so, I need to find $X$ so $A/Bcong A/Ccong X$. At start I though that $X=mathbb{Z}_2$ but I can't to seem to prove that $A/Bcong X$. How to find $X$?
abstract-algebra group-theory
abstract-algebra group-theory
asked Jan 12 at 18:14
abuka123abuka123
273
asked Jan 12 at 18:14
abuka123abuka123
273
asked Jan 12 at 18:14
abuka123abuka123
273
asked Jan 12 at 18:14
abuka123abuka123
273
asked Jan 12 at 18:14
abuka123abuka123
273
273
$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15
add a comment |
$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15
$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15
add a comment |
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$begingroup$
Don't $A/B$ and $A/C$ both have order $2$?
$endgroup$
– Lord Shark the Unknown
Jan 12 at 18:16
$begingroup$
@LordSharktheUnknown I think they do. then $X=mathbb{Z}_2$?
$endgroup$
– abuka123
Jan 12 at 18:18
$begingroup$
@abuka123: Not equal; just isomorphic, because that is the only group of order 2 (up to isomorphism).
$endgroup$
– Arturo Magidin
Jan 13 at 1:15