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I would like to find a condition for an exact sequence of abelian groups $$ 0\to H\to G\to K\to 0 $$ to split. Assume for simplicity that $H=\langle h \rangle$ is cyclic, and choose a basis for $G= \langle g_1 \rangle \oplus \ldots \oplus \langle g_n \rangle$. Write $h= \sum a_i g_i$, with $0 \leq a_i < o(g_i)$, where $o(g_i)$ is the order of $g_i \in G$.

By looking at examples, it seems to me that the exact sequence splits if and only if $$ \gcd(o(h), a_1, \ldots, a_n)=1. $$ Is this last statement correct? (and, if so, why?)

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3 Answers 3

Assuming we have the short exact sequence

$$0\longrightarrow H\stackrel{f}\longrightarrow G\stackrel{g}\longrightarrow K\longrightarrow 0$$

then the sequence splits iff there exists a homomorphism $\,\gamma:K\longrightarrow G\,$ s.t. $\,g\circ \gamma=Id_K\,$ , iff there exists a homomorphism $\,\phi:G\longrightarrow H\,\,s.t.\,\,\phi\circ f=Id_H\,$ .

You can find a proof of the above, sometimes known as the splitting lemma, here

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Thanks for your answer. Sorry but this looks to me just as the definition. Is it obvious that any of the three equivalent conditions proves (or disproves) the statement in my question? –  calc Sep 18 '12 at 21:29

Consider the short exact sequence

$$ 0\to H\stackrel{f}\to G\to K\to 0.$$

Let $H=\langle h\rangle$, and $f(h)=\displaystyle\sum_{i=0}^n a_i g_i$. Then the SES splits if and only if there exists a homomorphism $\phi\ :\ G\to H$, such that $\phi\circ f\equiv id$. Now if $\phi(g_i)=k_i h$, then $\phi$ is a homomorphism if and only if

$$ k_i |g_i|\equiv 0\pmod{|h|},\qquad 1\le i\le n.\qquad (1)$$

And we have $\phi\circ f\equiv id$ if and only if

$$ \displaystyle\sum_{i=1}^n a_i k_i\equiv 1\pmod{|h|}.\qquad (2)$$

Condition $(2)$ clearly implies the GCD condition is necessary. I think some Chinese Remainder Theorem stuff will show it is sufficient as well.

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Thanks! Yes, it seems to me that Chinese tells that GCD is equivalent to the existence of $k_i$ satisfying (1) and (2). –  calc Sep 19 '12 at 9:10
up vote 1 down vote accepted

This is wrong, as shown by the accepted answer to my question on MO: http://mathoverflow.net/questions/107768/on-the-existence-of-a-direct-summand-containing-a-fixed-subgroup

A simple example is the subgroup generated by $\langle (2,1) \rangle $ inside $\mathbb{Z}_8 \oplus \mathbb{Z}_2$. This is a subgroup of order $4$ whose quotient is cyclic of order $4$, generated by the equivalence class of $(1,0)$. Clearly $\mathbb{Z}_4 \oplus \mathbb{Z}_4 \neq \mathbb{Z}_8 \oplus \mathbb{Z}_2$.

The answer of Steve D shows that the GCD condition is necessary, but in fact it is not sufficient as this example clearly shows.

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