$\Bbb K$-algebra structure on isomorphic vector space Let $V,W$ be two isomorphic vector spaces over a field $\Bbb K$.
Suppose we find an operation $\star_V$ which makes $V$ a $\Bbb K$-algebra. Then I think even $W$ get the $\Bbb K$-algebra structure, under the isomorphism above.
Namely: let $\varphi:W\stackrel{\sim}\to V$ be the isomorphism, then we can define an operation $\star_W$ on $W$ in the obviuos way: given by $w_1\star_W w_2:=\varphi^{-1}(\varphi(v_1)\star_V\varphi(v_2))$, where $v_1,v_2$ are the corresponding elements in $V$ of $w_1,w_2$ resp. under the isomorphism $\varphi$.
Am I right? Or there exists some pathological situation? Maybe when $V,W$ are infinite dimensional vect.sp.
Thank you all
 A: Probably unrelated (so I don't want an upvote, but I won't miss the opportunity to tell you this).
This is the way you "transport" structures along sets with the same number of elements, provided one of them is endowed with some additional structure. And one of the byproducts of this procedure is that some collections you can consider in Mathematics are "too big to be sets".
You have to accept the fact that the collection $\mho$ of all sets is not a set: if it was, then you could consider its powerset $P(\mho)$; now $P(\mho)\in \mho$, a blatant contradiction (I'm being sloppy here).
Given this, I can prove that the collection $\mho_{\text{AbGrp}}$ of all abelian groups is not a set: it suffices to see that every set has at least a group structure. This is precisely where I transport structure along bijections.


*

*Any finite set has a group structure: easy peasy. If $|X|=n$ then you can transport structure along the bijection $X\to \mathbb{Z}/n$.

*Any countable set has a group structure: use $\mathbb{Z}$ (cyclic groups often save the day!).

*Any more-than-countable set $X$ has a group structure: let $\alpha=|X|$, notice that $\bigoplus_{i\in\alpha}\mathbb Z$ (direct sum of $\alpha$ copies of Z) is an abelian group and that there is a bijection $\alpha = \left|\bigoplus_{i\in\alpha}\mathbb Z\right|$.

