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I want to prove that the group ring $\mathbb{Z}_2[G]$ is not semisimple when $G$ is the cyclic group of order two. I guess the simplest way would be to show that there exists a submodule that is not a direct summand. However I do not know exactly how I should go about showing this.

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2 Answers 2

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The group ring $\mathbb{Z}_2[G]$ has only four elements: $0$, $1$, $x$ and $1+x$. A proper submodule $M$ will have order $2$; if $a+bx\in M$, then so is $x(a+bx) = b+ax$; so if $M=\{0, a+bx\}$, with $a$ and $b$ not both zero, then $b+ax=0$ (so $a=b=0$) or $b+ax=a+bx$ (so $a=b=1$). Thus, the only proper submodule is $\{0,1+x\}$, and it has no complement (since there are no other proper submodules).

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Is $\mathbb{Z}_2$ the field with two elements? If so, there's a non-split exact sequence

$$ 0 \to \mathbb{Z}_2 \to \mathbb{Z}_2G \to \mathbb{Z}_2 \to 0 $$

where the first map sends 1 to $1+g$ ($G = \langle g \rangle$) and the second is the augmentation map.

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