Why does Zorn's Lemma fail to produce a largest group?

Zorn's Lemma states that if every chain $\mathcal{C}$ in a partially ordered set $\mathcal{S}$ has an upper bound in $\mathcal{S}$, then there is at least one maximal element in $\mathcal{S}$.

Why can't we apply Zorn's Lemma in the following case?

Let $\mathcal{S}$ be the set of all groups. Define a partial order $\preceq$ as follows: for $H, G \in \mathcal{S}$, define $H \prec G$ if and only if $H$ is a subgroup of $G$. Then every chain $\mathcal{C}=(G_{\alpha})_{\alpha \in A}$ in $\mathcal{S}$ has an upper bound $\cup_{\alpha \in A} G_{\alpha}$ in $\mathcal{S}$. But certainly there is no maximal element in $\mathcal{S}$.

Could anyone tell me what is wrong with this argument?

There is no set of all groups, but Zorn's Lemma can be phrased for partially ordered classes as well, but we need to require that any chain has an upper bound, also proper class chains. However in that case it is easy to define a proper class chain which has no upper bound.

Simply take for each ordinal $\alpha$ the free group generated by $\alpha$. There are natural injections given by the natural injections between two ordinals. However this chain does not have an upper bound, since an upper bound would be a group, which is a set, that embeds all ordinals. This is a contradiction to Hartogs theorem, of course.

(You can think about this in the following analog: an infinite chain of finite sets, or finite groups, will not have an upper bound which is also finite.)

Because there is no such thing as “the set of all groups”.