The notions of $K$-finite vectors in admissible representations Let $G$ be a real Lie group and $K$ be a maximal compact subgroup. Let $(\pi, V)$ be a Hilbert space continuous representation of $G$. It can be restricted to $K$ as a representation of $K$. For each irreducible representation $\sigma$ of $K$, let $V(\sigma)$ be the $\sigma$-isotypic subspace of $V$, i.e. the sum over all (finite dimensional) subrepresentations of $V$ that are equivalent to $\sigma$.
Let $V_{\mathrm{fin}}$ be the algebraic direct sum $V_{\mathrm{fin}} := \oplus_{[\sigma] \in \widehat{K}} V(\sigma)$, where $\widehat{K}$ is the collection of irreducible representations of $K$. An element of $V_{\mathrm{fin}}$ is called a $K$-finite vector.
The Lemma 4.4.4 of the book by Getz and Hahn is left to the reader to prove:

Lemma 4.4.4: Let $\mathfrak{t} = \mathrm{Lie}(K)$, then the following statements are equivalent:
(1) The vector $\phi \in V$ is $K$-finite,
(2) The space $\langle \pi(k)\phi : k \in K \rangle$ is finite
dimensional.
Moreover, if $\phi$ is smooth, then this is equivalent to
(3) The space $\langle \pi(X)\phi : X \in \mathfrak{t} \rangle$ is
finite dimensional.

In the book, I was told to confer Proposition 2.4.4 of Bump's book Aut. forms and rep'ns.
However, in Bump's proposition, $(\pi, V)$ is assumed to be admissible and the groups $G$
is the specific groups $GL(2, \mathbb{R})$. So when trying to generalize to the above case, I met the following questions:


In the proof that (1) implies (2), for a $K$-finite vector $\phi$, it lies in the direct sum of a finite number of $V(\sigma)$, each of which is a finite-dimensional $K$-invariant subspace and the space spanned by the $K$-translates of $\phi$ is contained in this space and then (2) follows.

My question 1: In the above proof, the bolded part requires the finite-dimensionality of all $V(\sigma)$. It seems that this requires the admissibility of $V$? But the shaded quoted Lemma 4.4.4 do NOT require this condition. But I wonder how to get rid of the admissibiliy condition? (From the above proof, the "admissible" condition seems to be essential. So I doubt that whether we can really get rid of the condition.)

When reading the proof of (3) implying (2), the proof in Bump seems to rely highly on that $G=GL(2, \mathbb{R})$, using the explicit expression of the exponential map $\exp: \mathfrak{t} \rightarrow K$ as taylor series of matrices. Discussions are in the comments of $K$-finite vector equivalent to being $\operatorname{Lie}(K)$-finite .
Question 2: How to get rid of the assumption that $G$ is the general linear group?
Question 3: Actually since the post $K$-finite vector equivalent to being $\operatorname{Lie}(K)$-finite recieved no answer, could anyone answer that post or explain what was going on there?'

Sorry for such a long post and possiblly trivial questions as above as I am a new learner on automorphic representations. Thank you all for commenting and answering! :)
 A: It's certainly fair to wonder about the interdependencies among these adjectives! :)
First, even just for (locally compact, Hausdorff) topological groups $G$ and compact subgroups $K$ (a configuration that has proven useful to contemplate), there is the obvious possibility that a continuous repn of $G$ on a topological vector space $V$ is or is not "admissible", where admissible means that a given irreducible of $K$ occurs with finite multiplicity. This property already admits a few varying "definitions". Often, otherwise-nice, natural Hilbert space repns are not admissible: the space of $L^2$ level-one waveforms for $SL_2(\mathbb Z)$ on $\mathfrak H$ is infinite-dimensional (as Selberg showed... it's not obvious).
A sometimes-useful point is that even when a repn is not admissible, it may have $K$-finite vectors. And, even if the repn is admissible, not every vector need be $K$-finite: consider $G=K=circle$ and $L^2$ of the circle... it's surely admissible, but only finite Fourier series are $K$-finite.
Coming to the comparison of the Lie-algebra version of $K$-finiteness and the group-theoretic: one direction of the implication is clear, that $K$-finiteness implies $\mathfrak k$-finiteness. We could imagine that the opposite direction is merely technically more difficult, depending on finiteness of the number of connected components of $K$, but it's not quite so simple. That is, there are non-analytic but smooth functions. Yes, Harish-Chandra proved that (in most situations of interest) analytic vectors of a repn are dense...
But/and in finite-dimensional repns (such as a $K$-isotype in an admissible repn of $G$), all vectors are analytic... (and then the rest is just squabbling over definitions...)
One consolation is that very often it's not the provable equivalence of various related definitions that's relevant, but just the use of the words as a description of what's happening. :)
