Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
It seems that standard proof for commutative case works here as well and gives $AC = CA = E$. –  user87690 Sep 17 '13 at 17:51

1 Answer 1

up vote 3 down vote accepted

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 $x = x\lambda\pi(x)\cdot\lambda\pi(x^{-1})$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.