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I have a basic question from Rotman's 'An Introduction to Homological Algebra', Thm 5.3 pp 152.

Let \begin{equation*} 0\xrightarrow{} A\xrightarrow{} E\xrightarrow{\pi} G\xrightarrow{} 1 \end{equation*} be an exact sequence of groups where $A$ is an abelian normal subgroup of $E$ which is not necessarily abelian.

Suppose that the sequence is split on the right by a homomorphism $\lambda:G\rightarrow E$, i.e. $\pi\lambda=1_G$. Let $C=\text{im}\lambda\cong G$.

Question: How can we show that $A+C=E$?

Many thanks in advance.

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    $\begingroup$ It seems that standard proof for commutative case works here as well and gives $AC = CA = E$. $\endgroup$
    – user87690
    Sep 17, 2013 at 17:51

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If $E$ is not abelian you shouldn't write $A + C = E$, you should write $AC = E$ instead. Additive notation is generally reserved for abelian groups only.

Anyway, for $AC = E$ simply note that if $x \in E$ then $x = x\lambda\pi(x)\cdot\lambda\pi(x^{-1})$.

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