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We know two finite groups with the same character table might not be isomorphic (e.g. $D_4$ and $Q_8$), but the sizes of their abelianizations are equal (in fact equal to the number of linear characters, which can be read off the character table).

Can we also say the abelianizations of these groups are isomorphic, or is this not necessarily the case?

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Yes. The $1$-dimensional characters form a group under pointwise multiplication, and the abelianization is (noncanonically) isomorphic to this group. (It is canonically isomorphic to the Pontryagin dual of this group.)

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  • $\begingroup$ I think I can see how the 1-dimensional characters form a group under pointwise multiplication, but how is the abelianization isomorphic to this group? $\endgroup$ – Richard Jan 5 '16 at 3:04
  • $\begingroup$ @Richard: a $1$-dimensional character is a homomorphism $G \to \mathbb{C}^{\times}$, or equivalently a homomorphism $G/[G, G] \to \mathbb{Q}/\mathbb{Z}$ (since $G$ is finite, the image of a character lands in roots of unity). If $A$ is a finite abelian group, then the group of homomorphisms $A \to \mathbb{Q}/\mathbb{Z}$ (which is one possible definition of the Pontryagin dual) is (noncanonically) isomorphic to $A$. This follows from the structure theorem for finite abelian groups, together with the observation that this is true if $A$ is cyclic. $\endgroup$ – Qiaochu Yuan Jan 5 '16 at 3:09

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