Let $C=\langle c \rangle$ denote the cyclic group of order $2$ written multiplicatively, i.e. $C$ is a copy of $\{\pm 1\}$ with the usual product, and let $k$ be a field of characteristic different from $2$. Now consider the group algebra
$$k[C]=\{\alpha +\beta c|\alpha,\beta\in k\}$$
and the augmentation morphism
$$\lambda (\alpha +\beta c)=\alpha +\beta,$$
which has $\ker\lambda =k(1-c)$. We know that this ideal is actually the Jacbson radical of $k[C]$ and so that $\mathcal{U}(k[C])=k[C]\setminus \ker\lambda$, where $\mathcal{U}$ denotes the group of unit of a ring.
Now my problem comes when I consider the element $1+c$, which is not in $\ker\lambda$ since we are assuming $k$ has characteristic different from $2$, which means that it should be invertible. However $(1-c)(1+c)=1-c^2=0$, hence $1+c$ is a zero-divisor and it cannot be invertible!
Where am I wrong with my reasoning? I know it's something really trivial but I'm lost; any help is greatly appreciated.
If you calculate the regular matrix representation, $c$ is clearly represented by the matrix $\left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, which gives for $c+1$ the non-invertible matrix $\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$. So indeed $1+c$ is non-invertible. The statement that every element outside the Jacobson radical is invertible is wrong - it would imply, for example that every element of a semisimple algebra is invertible, which is patently untrue.