Hopf–Maschke theorem by integral averaging

Suppose $H$ is semisimple. By [F4]–[F5], write its left regular module as a direct sum of simple submodules, which are left ideals by [F8]. Only finitely many summands can be nonzero: if $n={\dim }_{k}H$ and there were more than $n$ nonzero summands, choosing one nonzero vector from each of $n+1$ distinct summands would give $n+1$ linearly independent vectors in $H$. Thus $H$ is a finite direct sum of simple modules.

Conversely suppose a left integral $t$ has $\epsilon (t)\mathrm{\ne }0$ and put $\mathrm{\Lambda }=t\mathrm{/}\epsilon (t)$. Then $\mathrm{\Lambda }$ is a left integral and $\epsilon (\mathrm{\Lambda })=1$. For every $h\in H$, $\sum h{\mathrm{\Lambda }}_1\otimes S({\mathrm{\Lambda }}_2)=\sum h_1{\mathrm{\Lambda }}_1\otimes S({\mathrm{\Lambda }}_2)S(h_2)h_3=\sum h_1{\mathrm{\Lambda }}_1\otimes S(h_2{\mathrm{\Lambda }}_2)h_3=\sum (h_1\mathrm{\Lambda }{)}_1\otimes S((h_1\mathrm{\Lambda }{)}_2)h_2=\sum {\mathrm{\Lambda }}_1\otimes S({\mathrm{\Lambda }}_2)h$. The first equality inserts $\sum S(h_2)h_3=\epsilon (h_2)1$; the second uses anti-multiplicativity [F7]; the third uses multiplicativity of $\mathrm{\Delta }$; the fourth uses $h_1\mathrm{\Lambda }=\epsilon (h_1)\mathrm{\Lambda }$.

The kernel $J=\ker \epsilon$ is a left ideal (in the sense of [F8]) since $\epsilon$ is an algebra map. By [F6], choose a left $H$-submodule $C$ with $H=J\oplus C$. The restriction $\epsilon {\mathrm{\mid }}_C:C\to k$ is injective because $C\cap J=0$, and is surjective because $\epsilon (H)=k$ and $\epsilon (J)=0$. Its inverse $s:k\to C\subset H$ is $H$-linear. Set $\mathrm{\Lambda }=s(1)$; then $h\mathrm{\Lambda }=s(h\cdot 1)=\epsilon (h)\mathrm{\Lambda }$ and $\epsilon (\mathrm{\Lambda })=1$.

Let $J$ be any left ideal of $H$, meaning a submodule of the regular left module as in [F8]. Fix a finite basis $e_1,\dots ,e_n$ of $H$. Start with an empty list and append a vector of $J$ outside the current span while one exists; this process stops after at most $n$ choices and yields a basis of $J$. Next append each $e_i$ that is outside the current span. This gives a basis of $H$ extending the basis of $J$, so projection onto $J$ along the span of the appended vectors gives a $k$-linear map $p:H\to J$ with $p{\mathrm{\mid }}_J={\mathrm{i}\mathrm{d}}_J$. Define $P(v)=\sum {\mathrm{\Lambda }}_{1}p(S({\mathrm{\Lambda }}_2)v)$. The values lie in $J$ because $p$ takes values in $J$ and $J$ is a left ideal. The tensor identity in step 1.2 gives $P(hv)=\sum {\mathrm{\Lambda }}_{1}p(S({\mathrm{\Lambda }}_2)hv)=\sum h{\mathrm{\Lambda }}_{1}p(S({\mathrm{\Lambda }}_2)v)=hP(v)$, so $P$ is $H$-linear. For $v\in J$, $S({\mathrm{\Lambda }}_2)v\in J$, whence $P(v)=\sum {\mathrm{\Lambda }}_{1}S({\mathrm{\Lambda }}_2)v=\epsilon (\mathrm{\Lambda })v=v$ by [F1]. Thus $P$ is a retraction onto $J$, and $H=J\oplus \ker P$.

Every submodule of the regular module $H$ is a left ideal by [F8] and therefore has a complement by step 2.2. Induct on ${\dim }_{k}H$: a nonzero finite-dimensional module has a nonzero submodule $U$ of least positive dimension (among the finitely many possible dimensions); it is simple because any proper nonzero submodule would have smaller positive dimension. Split $H=U\oplus M$. If $L\le M$, step 2.2 gives $H=L\oplus C$ for some submodule $C$; for any $m\in M$, write $m=l+c$. Since $l\in L\le M$, $c=m-l$ lies in $C\cap M$, and $L\cap (C\cap M)=0$, so $M=L\oplus (C\cap M)$. Thus every submodule of $M$ has a complement in $M$. Induction makes $M$ a finite direct sum of simple submodules, so $H=U\oplus M$ is semisimple. Hence $H$ is a semisimple ring by [F5]. The forward and reverse implications prove the equivalence.