In general, no, we need not have this situation.
For example, take the empty language and let $T=\{\forall x,y,z(x=y\vee x=z\vee y=z)\}$. Up to isomorphism there are exactly two elements of $T$, namely the one-element set and the two-element set (keep in mind that structures in the empty language are just sets). Up to $T$-provable equivalence, then, there are only four distinct sentences: the provable sentence $\top$, the disprovable sentence $\perp$, and the independent sentences $\exists x,y(x\not=y)$ and $\forall x,y(x=y)$. To convince yourself of this, note that we can think of a sentence up to logical equivalence as just its collection of isomorphism types of models, and since there are only two models of $T$ up to isomorphism there are only four collections of models of $T$ up to isomorphism.
However, this is fairly artificial. For theories like $\mathsf{ZFC}$ - that is, theories which let us "implement mathematics" - we do indeed always have weaker independent sentences:
If $\varphi$ is independent of $\mathsf{ZFC}$ then there is some $\theta$ which is also independent of $\mathsf{ZFC}$ such that $\mathsf{ZFC\cup\{\varphi\}}\vdash\theta$ but $\mathsf{ZFC\cup\{\theta\}\not\vdash\varphi}$.
Proof: Fix a sentence $\varphi$ which is independent of $\mathsf{ZFC}$ and consider the new theory $T=\mathsf{ZFC}\cup\{\neg\varphi\}$. By Godel's incompleteness theorem, $T$ must be incomplete; let $\eta$ be independent of $T$. But now consider the sentence $\eta\rightarrow\varphi$. Obviously $\mathsf{ZFC}\cup\{\varphi\}\vdash\eta\rightarrow\varphi$, so we just need to show $\mathsf{ZFC}\cup\{\eta\rightarrow\varphi\}\not\vdash\varphi$.
To do this, first note that $T\cup\{\neg\eta\}$ is consistent since $\eta$ is independent of $T$. In particular, this means that that $T\cup\{\neg\eta\}\not\vdash\varphi$, since $\neg\varphi\in T$. But we also have $\mathsf{ZFC}\subseteq T$ (by definition) and $\{\neg\eta\}\vdash\eta\rightarrow\varphi$ (since an implication with false hypothesis is true), so a fortiori we have $\mathsf{ZFC}\cup\{\eta\rightarrow\varphi\}\not\vdash\varphi$.
So as desired, $\eta\rightarrow\varphi$ is strictly weaker than $\varphi$. And we've used nothing about $\mathsf{ZFC}$ other than that it is essentially incomplete (= every computably axiomatizable theory containing it is incomplete), so the same argument applies to (first-order) Peano arithmetic and all the other theories subject to Godel.
Incidentally, it may be helpful to frame things algebraically. Given a theory $T$, for sentences $\varphi,\psi$ in the language of $T$ write "$\varphi\trianglelefteq\psi$" iff $T\vdash\varphi\rightarrow\psi$ (equivalently: iff $T\cup\{\varphi\}\vdash\psi$) and write "$\varphi\equiv\psi$" iff $T\vdash\varphi\leftrightarrow\psi$.
For example:
Every Lindenbaum algebra is a Boolean algebra (at least in classical logic), with Boolean structure corresponding to the logical connectives: least upper bounds are given by $\vee$, greatest lower bounds by $\wedge$, complementation by $\neg$, and the top and bottom elements are $\top$ and $\perp$ respectively.
- Note that weaker sentences are higher in the Lindenbaum algebra; this may be counterintuitive at first, but it's ultimately quite convenient.
An inconsistent theory has a one-element Lindenbaum algebra. A complete consistent theory has a two-element Lindenbaum algebra. If $T\subseteq S$, then the Lindenbaum algebra of $S$ is a quotient of the Lindenbaum algebra of $T$ (adding more axioms makes the theory bigger but the algebra smaller).
Any Boolean algebra with no coatoms (maximal non-$1$ elements) also has no atoms (minimal non-$0$ elements). Essential incompleteness of a theory implies the nonexistence of coatoms, hence for purely algebraic reasons the nonexistence of atoms; so in a precise sense, the proof above that $\mathsf{ZFC}$ has no "weakest possible independent sentences" was purely algebraic (Godel's theorem says no coatoms, so we in turn get no atoms).
More generally, thinking about the complement operation in a Boolean algebra we have that when a given property holds of a given Lindenbaum algebra, so does its "dual property" (e.g. "No coatoms" vs. "No atoms," "Arbitrary joins exist" vs. "Arbitrary meets exist," etc.).
This algebraic approach to logic is super useful; it shows up in model theory in a "geometric" guise as the space of types, and it's also the motivating idea for the subject of algebraic logic.
Best Answer
For any (finite or infinite) cardinal $\kappa$, if $\kappa \ge 5$ then the finitary alternating group $A(\kappa)$ is simple. This is the group consisting of permutations of $\kappa$ that have finite support and are even. (If $\kappa$ is finite then this is just the ordinary alternating group $A_\kappa$.) If $\kappa$ is infinite, then $A(\kappa)$ has cardinality $\kappa$.