Theorem 5.11 in Hungerford Algebra $f^{-1}f(K) = K$ if and only if $\ker f < K$.

Question

Let $f \colon G \to H$ be an homomorphism, in the proof of the theorem 5.11, Hungerford states that $f^{-1}(f(K)) = K$ if and only if $\ker f < K$ for $K$ is a subgroup of $G$.

I proved the forward direction which is rather straightforward. But Im struggling with the backward direction.

Suppose $\ker < K$, how to prove that $f(f^{-1}(K)) = K$?

well, maybe you mean $\dim \ker f < \dim K$ ("<" is used between numbers, "$\subset$" between sets) — the_candyman, May 14 '21 at 08:16
@the_candyman This is just the notations from the Hungerford....he uses $H < G$ to denotes subgroups. I thought its pretty common to use $H \leq G$ for subgroups, both Hungerford and Dummit&foote do this. — Andy, May 14 '21 at 08:19
More generally, $f^{-1}(f(K)) = \langle K, \ker f \rangle $. See https://math.stackexchange.com/a/955413/589 — lhf, May 20 '21 at 21:13

score 0 · Answer 1 · answered May 14 '21 at 09:09

0

Hint:

Show $f^{-1}\bigl(f(K)\bigr)=K\,\ker f$ (in multiplicative notation), and use the $2^{\textit{nd}}$ isomorphism theorem.

answered May 14 '21 at 09:09

Bernard

179,256

score 0 · Accepted Answer · answered May 20 '21 at 18:04

Elementary proof.

You already proved that if $f^{-1}(f(K)) = K$, then $\ker f \subseteq K$. Suppose now that $\ker f \subseteq K$ and let $x \in f^{-1}(f(K))$. Then $f(x) \in f(K)$ and hence there exists $y \in K$ such that $f(x) = f(y)$. It follows that $f(xy^{-1}) = f(x)f(y)^{-1} = 1$ and thus $xy^{-1} \in \ker f$, whence $xy^{-1} \in K$. Since $y \in K$, it follows that $x = (xy^{-1})y \in K$. Thus $f^{-1}(f(K)) \subseteq K$. The opposite inclusion $K \subseteq f^{-1}(f(K))$ is trivial, and thus $f^{-1}(f(K)) = K$.

Theorem 5.11 in Hungerford Algebra $f^{-1}f(K) = K$ if and only if $\ker f < K$.

2 Answers2