Unboundedness of the set of subgroups of a cardinal

Question

I'm trying to figure out why the following is true:

Let $ \kappa $ be an uncountable, regular cardinal. Suppose we turn it into a group (i.e. there are operations $ (\cdot, ^{-1}, e) $ with which $ \kappa $ is a group. My aim is to prove that the set

$$ \{ \alpha \in \kappa : \alpha \text{ is a subgroup of } \kappa\}$$

is a club. It's rather obvious that it is closed, however I'm not quite sure why it would be unbounded. Suppose we have an $ \alpha \in \kappa $ which is a subgroup. Somehow we need to enlarge $ \alpha $ so that it will still be both a group and an ordinal.

We can close $ \alpha + 1 $ under the group operations, but we have no warranty that it will be an ordinal (it wont be equal to $ \kappa $ since $ \kappa $ is regular).

I don't see any way to approach this. I would appreciate a hints

Answer 1

HINT: Suppose that $\alpha$ is a subgroup of $\kappa$. Let $G_0$ be the closure of $\alpha+1$ under the group operations. Let $\beta_0=\sup G_0$, and let $G_1$ be the closure of $\beta_0+1$ under the group operations. In general, given a bounded subgroup $G_n$ of $\kappa$, let $\beta_n=\sup G_n$, and let $G_{n+1}$ be the closure of $\beta_n+1$ under the group operations. Can you finish from here?

Answer 2

Take a ordinal $\alpha_0$. Let $G(\alpha_0)$ be the group generated by $\alpha_0$. Then $|G(\alpha_0)|<\kappa$. Define $\alpha_1$ to be any ordinal larger than all element of $G(\alpha_0)$ repeat this construction with $\alpha_1$. This gives $\alpha_n$ increasing. Your desired ordinal subgroup is then $\alpha=\sup\alpha_n$.