# Perfect core is homomorph-containing

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., perfect core) always satisfies a particular subgroup property (i.e., homomorph-containing subgroup)}
View subgroup property satisfactions for subgroup-defining functions $|$ View subgroup property dissatisfactions for subgroup-defining functions

## Statement

Suppose $G$ is a group and $H$ is the perfect core of $G$: it is the unique largest perfect subgroup of $G$. Then, $H$ is a homomorph-containing subgroup of $G$: given any homomorphism $\varphi:H \to G$, we have $\varphi(H) \le H$.

## Facts used

1. Perfectness is quotient-closed: The image of a perfect group under a surjective homomorphism is perfect.

## Proof

Given: A group $G$ with perfect core $H$ and a homomorphism $\varphi:H \to G$.

To prove: $\varphi(H) \le H$.

Proof: $\varphi(H)$ is perfect by fact (1). By the definition of perfect core, any perfect subgroup of $G$ is contained in $H$. Thus, $\varphi(H) \le H$.