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I'm just trying to solve an exercise from "A course in homological algebra" by Hilton and Stammbach, however I couldn't find any example.

Find two short exact sequences of abelian groups

$0 \longrightarrow A'\longrightarrow A \longrightarrow A'' \longrightarrow 0$

$0 \longrightarrow B'\longrightarrow B \longrightarrow B'' \longrightarrow 0$

such that two of the abelian groups belonging to distinct sequences are isomorphic, however the third is not, e.g., $A \cong B$ and $A'\cong B'$, but $A''\ncong B''$.

I have already found an example using semi direct product, however the resulting group is not abelian, just pick $A = G\ltimes H$ and $B = G \bigoplus H$. I think that two subgroups $HK = HN$ of a group $G$ such that $K \neq N$ will suffice one of the cases, however I did not found any concrete example.

Thanks in advance.

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Simplest example... $A', A'', B', B''$ are all of order $2$, $A$ is cyclic, $B$ is Klein $4$-group. –  GEdgar Jan 12 '13 at 21:45
2  
"such that two of them are isomorphic" --- such that two of what are isomorphic? –  Gerry Myerson Jan 12 '13 at 21:46
    
@GerryMyerson By two of them, I mean two of the abelian groups such that one do not belong to the other sequence. For instance $A \cong B$ and $A'\cong B'$, but $A''\ncong B''$. –  user40276 Jan 12 '13 at 21:50
2  
$A=A'=B=B'=\mathbb Z$, $A''=\mathbb Z_2$ $B''=\mathbb Z_3$. The map $A'\to A$ is multiplication by $2$. The map $B'\to B$ is multiplication by $3$. –  Grumpy Parsnip Jan 12 '13 at 21:59
    
@JimConant In your example you used the trick that you cannot think about $A''= A / A'$ because the group is infinite and the homomorphism is not surjective despite of the cardinality of $A$ and $A'$ be the same. But, just by curiosity, is it possible to find an example using only finite groups? –  user40276 Jan 12 '13 at 22:22

3 Answers 3

Here is a simple one:

$0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0 \longrightarrow 0$

$0 \longrightarrow 2\mathbb{Z}\longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z}/2\mathbb{Z} \longrightarrow 0$

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I didn't notice Jim Conant had already given essentially this same example in the comments before I posted it... –  mbrown Jan 12 '13 at 22:57

Going by the comment, I think this works: let $A=B=C_2\oplus C_4$. Then there are subgroups $A'$ and $B'$, both isomorphic to $C_2$, but one has quotient group $C_4$, the other, $C_2\oplus C_2$.

I'm using $C_n$ for the cyclic group of order $n$.

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Sorry, but how did you get the $C_2 \bigoplus C_2$? –  user40276 Jan 12 '13 at 22:00
    
Let $A$ be generated by $a$ and $b$ subject to $a^2=1$, $b^4=1$, $ab=ba$, and let $B'$ be the subgroup generated by $b^2$. –  Gerry Myerson Jan 15 '13 at 5:07

I think one of the simplest examples is the following:

$$1\longrightarrow C_2\longrightarrow C_4\longrightarrow C_2\longrightarrow 1$$

$$1\longrightarrow C_2\longrightarrow C_2\times C_2\longrightarrow C_2\longrightarrow 1$$

$\,C_k=$the cyclic group of order $\,k\,$

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The OP asked about abelian groups. –  Grumpy Parsnip Jan 13 '13 at 1:43
1  
Thanks, I didn't notice that. I've changed the answer accordingly. –  DonAntonio Jan 13 '13 at 2:40

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