Suppose $\kappa$ is an uncountable regular cardinal, and $(\kappa, \cdot, ^{-1}, e$) is a group. Prove that that $C = \{\alpha < \kappa: \alpha\, \textrm{is a subgroup of}\, \kappa)$ is unbounded in $\kappa$.
It's easy to see that $C$ is actually closed: if $\alpha_\beta$ are subgroups of $\kappa$ for $\beta < \theta$, then $\bigcup_{\beta < \theta} \alpha_\beta$ is also a subgroup, as a union of increasing sequence of groups is again a group. I fail to see how to exhibit arbitrarily large ordinal subgroups in $\kappa$, though.
My problem is that if we take some $\alpha_0 < \kappa$ and close $\alpha$ under group operations, we'll get some subgroup $U_0$ of $\kappa$ with cardinality $< \kappa$. The thing is that $U_0$ is not necessarily ordinal, so we consider $\alpha_1 = \sup U_0$, close $\alpha_1$ under group operations to obtain some subgroup $U_1$ with $|U_1| < \kappa$, and then transfinitely continue this process until some $U_\beta$ will turn out to be ordinal. The problem is that we may get $\beta \geq \kappa$, and so $U_\beta = \kappa$ which is not a proper subgroup.
HINT: If $\alpha=\sup_{n\in\omega}\alpha_n$, and $F$ is a finite subset of $\alpha$, then $F\subseteq\alpha_n\subseteq U_n$ for some $n\in\omega$.