Subgroups of an infinite cardinal of uncountable cofinality. I am doing the following exercise and I cannot see what am I doing wrong (I'm not using the cofinality hypothesis).
Let $\kappa$ be an infinite cardinal of uncountable cofinality together with a group operation $\star$. I want to show that the set $C$ of $\alpha \in \kappa$ such that $(\alpha,\star)$ is a subgroup of $(\kappa,\star)$ is a club of $\kappa$ (closed unbounded subset).
That $C$ is closed is quite straightforward, let's prove that it is not bounded : let $\alpha \in \kappa$. We will define by recursion a $\gamma \in \kappa$ such that $\alpha \in \gamma$ and $\gamma$ is a subgroup of $\kappa$.
As $\kappa$ is a group for all $\nu \in \alpha$ there is a $\beta_{\nu}\in\kappa$ such that $\nu^{-1}\in\beta_{\nu}$. Let $\beta_0:= sup_{\nu\in\alpha}\beta_{\nu}$ and we have $\beta_0=\beta_0^{-1}$.
Let now take $\nu_1,\nu_2\in\beta_0$, we can find $\delta_{\nu_1,\nu_2}\in\kappa$ such that $\nu_1\star\nu_2\in\delta_{\nu_1,\nu_2}$ and we define $\delta_0:= sup_{(\nu_1,\nu_2)\in \beta_0^2}\delta_{\nu_1,\nu_2}$.
Defining the limit cases as the union of the previous $\beta$´s or $\delta$'s we construct a sequence:
$$\beta_0\in\delta_0\in\beta_1\in\delta_1\in...$$
and we define $\gamma:=sup_{\nu\in\kappa}\beta_{\nu}=sup_{\nu\in\kappa}\delta_{\nu}$. We have obtained a $\gamma$ such that $\kappa\ni\gamma \ni \alpha$ and $(\gamma,\star)$ is a group.
What am I missing here? 
 A: Here's the issues I see.
First, in the portion that begins "As $\kappa$ is a group for all $\nu \in \alpha$..." you seem to have a conflict of notation. By "$\nu^{-1}$", do you mean the inverse element of $\nu$, or the set of all inverses of elements of $\nu$? You seem to be using it the first way at first, but when you claim $\beta_0 = \beta_0^{-1}$ you have to be using it in the second way. Note: This is a purely notational problem; the reasoning is fine so far.
Second, $\beta_0$ certainly has all of the inverses of elements of $\alpha$ - but why does it have to contain the inverse of $\alpha + 1$? This isn't actually an issue - if you take a close look at your final step you'll see you don't need $\beta_0$ to be closed under inverses - but it sure looks like what you're claiming.
Third (and this part deals with cofinality): $\beta_0$, as you've defined it, is the supremum of a sequence of length $\alpha$. If $\mathrm{cf}\kappa < \alpha$, there's no guarantee $\beta_0$ will be less than $\kappa$!
Finally, another cofinality issue: your sequence of $\beta$s and $\delta$s has length $\omega$. To know that its limit is below $\kappa$, you need to have that $\kappa$ has uncountable cofinality.
Big Issue: As far as I can tell, the claim you're trying to prove is false. Consider the following group: let $\kappa = \aleph_{\omega_1}$, with the operation defined with $0$ the identity, $n^{-1} = \aleph_{n + 1}$ for finite $n$, and $\alpha^{-1} = \aleph_{\alpha}$ for infinite countable ordinals $\alpha$. Define your group operation in whatever way you like that's compatible with the above. Now no initial segment of $\kappa$ except $0$, $1$, and $\kappa$ is even closed under inverse, let alone a subgroup.
This problem would be fixed if $\kappa$ was taken to be regular, instead of just having uncountable cofinality, or if somehow inverses were assumed bounded (i.e., if the set of inverses of ordinals below $\alpha$ was guaranteed to be bounded below $\kappa$; this would be a constraint on the structure of the group). One appealing possibility is that both the inverse operation and the group operation might be required to be continuous. Are you sure that's not what you're being asked to prove?
