1
$\begingroup$

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).

$\endgroup$
  • 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
6
$\begingroup$

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 $\mathbb C$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.