# How to calculate inverses in the group algebra

Is there an algorithm to calculate the inverse of an element in the group algebra?

For example, does the element $(1 2 3) + 2 . (1 2)(3 4)$ in the group algebra $\mathbb{C} S_{4}$ have an inverse, and if so, how do I calculate it?

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Where (in which group) is all this happening? Anyway look at my answer and leave your comments as to how satisfied you're! I'll edit to add relevant information whenever needed. Explain your notation as well! –  user21436 Jan 23 '12 at 20:52
I'm after inverses in the group algebra, not in the group. In the example, the underlying group is S4, and the field might as well be C. So the group algebra is CS4, consisting of C-linear combinations of elements of S4. –  DavidA Jan 23 '12 at 20:57
My answer is vanishing! I am good at misreading things! Good that you didn't downvote! –  user21436 Jan 23 '12 at 20:58

In general, an element $u = \sum_{g \in G} \alpha_g g$ of the group algebra $\mathbb{C}G$ of a finite group $G$ is invertible if and only if $u\sigma$ is an invertible matrix whenever $\sigma$ is an irreducible matrix representation of $G$ (this only depends on the equivalence type of $\sigma$). This also gives a method to construct $u^{-1}$ explicitly when it exists, since the group algebra is a direct sum of matrix algebras, one for each equivalence type of irreducible matrix representation. For larger groups, this is probably impractical, at least by hand. In the particular example you choose, you can note that the element you consider is invertible if and only if $I + 2(321)(12)(24)$ is invertible ( I have multiplied by the inverse of the invertible element $(123))$. Since $(321)(12)(24)$ is a $3$-cycle and all $3$-cycles are conjugate, this element is invertible if and only if $v = I + 2(123)$ is invertible. Now for any choice of irreducible matrixrepresentation $\sigma,$ note that the eigenvalues of $v\sigma$ have the form $1+ 2 \omega$ for some third root of unity $\omega$ (possibly $1$). Such an expression is never zero, so $v$ is invertible. I omit the explicit calculation of the inverse, which is straightforward.