# Find two short exact sequences of abelian groups such that two of them are isomorphic, however the third is not

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.

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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
"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
$A=A'=B=B'=\mathbb Z$, $A''=\mathbb Z_2$ $B''=\mathbb Z_3$. 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

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
Thanks, I didn't notice that. I've changed the answer accordingly. –  DonAntonio Jan 13 '13 at 2:40