Is this statement true?

Let $G$ and $H$ be a group, and $G \cong H$. Then $\mathbb C G \cong \mathbb C H$?

For example I want to show that $\mathbb C G \cong \mathbb C H$ where

$$\begin{align*} G = ~ &\bigl\langle x,y\ \mid \ xyx=yxy\bigr\rangle\\ H = ~ &\bigl\langle a,b\ \mid \ a^2=b^3\bigr\rangle \end{align*}$$

Than I just need to show ($G \cong H$ which O know how to do).

  • 2
    $\begingroup$ Yes. Taking group algebras is a functor, and any functor sends isomorphisms to isomorphisms. $\endgroup$ – Qiaochu Yuan Sep 24 '15 at 21:06

Qiaochu's comment is right, but I don't want people to get the impression that one needs to know what a functor is in order to answer this question. If $f:G\to H$ is an isomorphism, then you can define an isomorphism $\hat f:\mathbb CG\to\mathbb CH$ by $$ \hat f\left(\sum_ic_ig_i\right)=\sum_ic_if(g_i) $$ for all coefficients $c_i\in\mathbb C$ and all group elements $g_i\in G$. It is straightforward to check that $\hat f$ is an isomorphism of algebras over $\mathbb C$.

  • $\begingroup$ That's what it means for taking group algebras to be a functor! $\endgroup$ – Qiaochu Yuan Sep 24 '15 at 21:25
  • 3
    $\begingroup$ @QiaochuYuan I almost agree; the only quibble is that I described a functor only on the subcategory of groups and isomorphisms (which is enough for this problem but not for a reasonable functorial view of group algebras). But the point of my answer is that people who've never heard of functors (and I conjecture the OP is such a person) can solve the problem without first learning what a functor is. With luck, someone might use my answer and your comment on it as a first step toward learning about functors. $\endgroup$ – Andreas Blass Sep 24 '15 at 21:31
  • $\begingroup$ Just a further question, the converse doesn't hold right? $\endgroup$ – SamC Sep 24 '15 at 21:33
  • $\begingroup$ @Andreas: your description of $\hat{f}$ makes no use of the fact that $f$ is an isomorphism! In any case, the reason my comment wasn't posted as an answer is that it wasn't an answer. I'm in no way saying that this answer was redundant. $\endgroup$ – Qiaochu Yuan Sep 24 '15 at 22:48
  • $\begingroup$ @SamC: yes. For example, if $G$ is a finite group, the isomorphism type of $\mathbb{C}[G]$ only knows the dimensions of the complex irreducible representations of $G$ (by Maschke's theorem + the Artin-Wedderburn theorem), so for example $\mathbb{C}[Q_8] \cong \mathbb{C}[D_4]$. $\endgroup$ – Qiaochu Yuan Sep 24 '15 at 22:49

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