I am reading the chapter on large cardinals in Jech's book and I found the following lemma:
If $\kappa \rightarrow (\kappa)^{<\omega} $ and if $\lambda < \kappa $ is a cardinal, then $ \kappa \rightarrow (\kappa)^{<\omega}_\lambda $.
Proof. Let $ F:[\kappa]^{<\omega} \rightarrow \lambda $ be a partition $ \lambda < \kappa $ pieces. We consider the following partition $G$ of $[\kappa]^{<\omega}$ into two pieces: If $\alpha_1 < . . . < \alpha_k < α_{k+1} <. . . < \alpha_{2k} $ are elements of $ \kappa$ and if $F({\alpha_1, . . . , \alpha_k}) = F({\alpha_{k+1}, . . . , \alpha_{2k}})$, then we let $G({\alpha_1, . . . , \alpha_{2k}}) = 1$; for all other $x \in [\kappa]^{<\omega}$, we let $G(x) = 0$. Now, let $H \subset \kappa$ be a homogeneous set for $G$, $|H| = \kappa$. We claim that for each $k$ and each $x \in [H]^{2k}$, $G(x) = 1$: This is because $|H| = \kappa > \lambda$, and therefore we can find $\alpha_1 < . . . < \alpha_k < \alpha_{k+1} < . . . < \alpha_{2k}$ in $H$ such that $F({\alpha_1, . . . , \alpha_k}) = F({\alpha_{k+1}, . . . , \alpha_{2k}})$.
My question is why the last argument is valid? I don't understand why $G$ gets $1$ on $H$ and not $0$ even when fixing the problem with definition of $G$ to be only on the even numbers.