For cardinals, if $\mathfrak{a}\ne\mathfrak{b}$ then $2^\mathfrak{a}\ne 2^\mathfrak{b}$

Question

In the usual ZF (or ZFC) set theory, let $\mathfrak{a}$ and $\mathfrak{b}$ be cardinal numbers. Is it correct that one can neither prove nor disprove the statement:

$$\mathfrak{a}\ne\mathfrak{b} \quad\Rightarrow\quad 2^\mathfrak{a}\ne 2^\mathfrak{b}$$

(clearly it is true for finite cardinals since $n\mapsto 2^n$ is injective as a map from $\mathbb{N}$).

Does the statement (distinct cardinals have distinct power-set cardinals) have a name? If we take this as an additional axiom to our set theory, what consequences does it have?

How is this related to other "extra" axioms? It appears to be provable from the (generalized?) continuum hypothesis (GCH)? But surely the GCH must be stronger.

It is correct that this cannot be proved or disproved from $\sf ZF(C)$. I don't know about an official name, I used "Injective Continuum Function" ($\sf ICF$). There has been a question about its consequences before, let me find one. — Asaf Karagila, Nov 21 '14 at 16:28
Also, yes, $\sf GCH\implies ICF$, and the other direction is false. — Asaf Karagila, Nov 21 '14 at 16:43
Great link. That gave me something to search for (I was not aware that $\mathfrak{a}\mapsto 2^\mathfrak{a}$ was called the continuum function). This highly related thread also turned up: Why continuum function isn't strictly increasing?. — Jeppe Stig Nielsen, Nov 21 '14 at 18:32
OK, I see now that really many people asked this question already. Will only link to Sets question, without Zorn's lemma (duplicate) also. But follow linked threads' linked threads. Maybe this question can simply be closed... — Jeppe Stig Nielsen, Nov 21 '14 at 18:41
It can be easily arranged to have this question closed as a duplicate. But the question is duplicate of what? you ask, essentially two questions here, each appeared in a separate thread. Let me know and I'll use my dupe-hammer to close it. — Asaf Karagila, Nov 21 '14 at 19:01

For cardinals, if $\mathfrak{a}\ne\mathfrak{b}$ then $2^\mathfrak{a}\ne 2^\mathfrak{b}$

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