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Suppose we have a short sequence of groups

\begin{equation*} 0\xrightarrow{} A\xrightarrow{} E\xrightarrow{\pi} G\xrightarrow{} 1. \end{equation*}

In some books, the authors mention that we can identify $A$ with a normal subgroup of $A$. Why?

I can only show that $A$ is isomorphic to the image of $A$, which can be viewed as a subgroup of $E$.

Shaun
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math112358
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1 Answers1

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Those authors are assuming that this is a short exact sequence of groups. Exactness means that the image of each map is exactly equal to the kernel of the next map. In particular, this means that the image of $A \to E$ is equal to the kernel of $E \to G$ and kernels are normal subgroups, hence the image of $A \to E$ is a normal subgroup.

Jim
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