Find a module $M$ and a submodule $N$ such that $|M| = 100$, $|N| = 20$, $M$ is not cyclic, and the exact sequence $0 \rightarrow M \rightarrow N \rightarrow M/N \rightarrow 0$ does not split.

I have tried $M = Z/2 \oplus Z/2 \oplus Z/5 \oplus Z/5$ and $N = Z/2 \oplus Z/2 \oplus Z/5$, but I don't know how to interpret $M/N$, let alone prove/disprove $M \cong M/N \oplus N$ (so that the sequence doesn't split).

Any help? (explanation would be much appreciated)

  • 1
    $\begingroup$ What ground ring? $\endgroup$ – Randall Dec 21 '17 at 17:06
  • $\begingroup$ Try putting a $Z/25$ in $M$. $\endgroup$ – Lord Shark the Unknown Dec 21 '17 at 17:09
  • $\begingroup$ @Randall whatever you want. $\endgroup$ – user484920 Dec 21 '17 at 18:17

We'll work with $\mathbb Z$-modules, i.e., abelian groups. Note $|M/N| = |M|/|N| = 5$ so $M/N$ is going to be cyclic of order $5$. Thus we can guarentee that the sequence does not split if $\mathbb Z/5$ is not a direct summand of $M$.

One group that has $\mathbb Z/5$ as a quotient but not as a summand is $\mathbb Z/25$ so this is a good candidate to build $M$ from. To get the right order lets choose $M = \mathbb Z/4 \oplus \mathbb Z/25$. A subgroup of order $20$ inside $M$ is $N = \mathbb Z/4 \oplus 5\mathbb Z/25$. We know just by order that $M/N \simeq \mathbb Z/5$ and we have chosen $M$ so that this isn't a summand. So our SES won't split.

If you want an explicit proof that it doesn't split note that $M/N$ is generated by $(0, 1) + N \in M/N$ so if there's a splitting then it comes from a homomorphism $\phi\colon\mathbb Z/5 \to \mathbb Z/4\oplus\mathbb Z/25$ satisfying $\phi(1) = (a, 1 + 5b)$ where $a \in \mathbb Z/4$ and $b \in \mathbb Z/25$. This means $\phi(0) = 5\phi(1)$ better equal $(0, 0)$ if such a homomorphism is possible. Show that $5\phi(1) \neq (0, 0)$.


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