Module is $\mathbb{F}_2[S_3]$-projective iff it's $\mathbb{F}_2[S_2]$-projective. Let $S_2\subset S_3$ be the subset of the symmetric group that fixes 3, i.e. $S_2=\{1, (12)\}$. Consider the group algebras $A:= \mathbb{F}_2[S_2]\hookrightarrow \mathbb{F}_2[S_3]=: B$. If $M$ is a (finitely generated) $B$-module, prove it's projective as a $B$-module iff it's projective as an $A$-module.
The $\Rightarrow$ direction is easy, because if $M$ is a direct summand of a free $B$-module, then it's easy to see $B\cong A^3$ as $A$-modules and so $M$ is a direct summand of a free $A$-module; unfortunately I'm stuck trying to prove the $\Leftarrow$ direction. Apologies for my very simple question, but unfortunately I've become blocked trying to understand the reason.
 A: Here is the proof I outlined in the comments. I will show a more general fact:

Theorem 1. Let $G$ be a finite group. Let $H$ be a subgroup of $G$. Let
  $K$ be a commutative ring such that $\left[  G:H\right]  $ is invertible in
  $K$. Let $M$ be a $K\left[  G\right]  $-module. Then, $M$ is projective as a
  $K\left[  G\right]  $-module if and only if the restriction
  $\operatorname{Res}_{H}^{G}M$ is projective as a $K\left[  H\right]
$-module.

(Here and in the following, "module" means "left module".)
Our proof of Theorem 1 will rely on the following neat fact:

Theorem 2. Let $G$ be a finite group. Let $H$ be a subgroup of $G$. Let
  $K$ be a commutative ring such that $\left[  G:H\right]  $ is invertible in
  $K$. Let $M$ be a $K\left[  G\right]  $-module. Then, $M$ appears as a direct
  addend in the $K\left[  G\right]  $-module $\operatorname{Ind}_{H}
^{G}\operatorname{Res}_{H}^{G}M$.

Here, if $V$ is any $K\left[  H\right]  $-module, then
$\operatorname{Ind}_{H}^{G}V$ is defined as the
$K\left[  G\right]  $-module $K\left[
G\right]  \otimes_{K\left[  H\right]  }V$, where $K\left[  G\right]  $ is
considered as a $\left(  K\left[  G\right]  ,K\left[  H\right]  \right)
$-bimodule (with both actions given by multiplications inside the group
algebra $K\left[  G\right]  $).
Our proof of Theorem 2 will rely on the following standard fact:

Lemma 3. Let $A$ be any ring. Let $U$ and $V$ be two $A$-modules. Let
  $P:U\rightarrow V$ and $Q:V\rightarrow U$ be two $A$-module homomorphisms
  satisfying $Q\circ P=\operatorname*{id}$. Then, $U$ appears as a direct addend
  in $V$.

Proof of Lemma 3 (sketched). The map $P$ is injective (since $Q\circ
P=\operatorname*{id}$); thus, $P\left(  U\right)  \cong U$ as $A$-modules. Now
argue that $V=P\left(  U\right)  \oplus\operatorname*{Ker}Q$ (internal direct
sum), since each $v\in V$ can be written as $v=\underbrace{P\left(  Q\left(
v\right)  \right)  }_{\in P\left(  U\right)  }+\underbrace{v-P\left(  Q\left(
v\right)  \right)  }_{\in\operatorname*{Ker}Q}$ (and since every $v\in
P\left(  U\right)  \cap\operatorname*{Ker}Q$ satisfies $v=0$ due to $Q\circ
P=\operatorname*{id}$). Thus, $P\left(  U\right)  $ is a direct addend of $V$.
Hence, $U$ appears as a direct addend in $V$ (since $P\left(  U\right)  \cong
U$). This proves Lemma 3. $\blacksquare$
Proof of Theorem 2. We shall denote the restriction
$\operatorname{Res}_{H}^{G}M$ as $M$ when the context makes it clear
that we mean a
$K\left[  H\right]  $-module (and not a $K\left[  G\right]  $-module). This is
harmless, since $\operatorname{Res}_{H}^{G}M=M$ as $K$-modules.
The definition of
$\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$ yields
\begin{equation}
\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}
^{G}M=K\left[  G\right]  \otimes_{K\left[  H\right]  }
\underbrace{\operatorname{Res}_{H}^{G}M}_{=M}=K\left[  G\right]
\otimes_{K\left[  H\right]  }M.
\end{equation}
For each $g\in G$, we define a $K$-linear map
\begin{align*}
Z_{g}:M  &  \rightarrow K\left[  G\right]  \otimes_{K\left[  H\right]  }M,\\
m  &  \mapsto g\otimes_{K\left[  H\right]  }g^{-1}m.
\end{align*}
For each $g\in G$, we consider the left coset $gH\in G/H$. It is easy to see
that if two elements $u$ and $v$ of $G$ belong to the same left coset
$\alpha\in G/H$, then
\begin{equation}
Z_{u}=Z_{v}.
\label{darij1.pf.t2.3}
\tag{1}
\end{equation}
[Proof of \eqref{darij1.pf.t2.3}: Let $u$ and $v$ be two elements of $G$
that belong to the same left coset $\alpha\in G/H$. Thus, $u\in\alpha$ and
$v\in\alpha$. From $v\in\alpha$, we obtain $\alpha=vH$ (since $\alpha$ is a
left coset of $H$), so that $u\in\alpha=vH$. In other words, $u=vh$ for some
$h\in H$. Now, for each $m\in M$, we have
\begin{align*}
Z_{u}\left(  m\right)   &  =u\otimes_{K\left[  H\right]  }u^{-1}m\qquad\left(
\text{by the definition of }Z_{u}\right) \\
&  =vh\otimes_{K\left[  H\right]  }\underbrace{\left(  vh\right)  ^{-1}
}_{=h^{-1}v^{-1}}m\qquad\left(  \text{since }u=vh\right) \\
&  =vh\otimes_{K\left[  H\right]  }h^{-1}v^{-1}m=v\otimes_{K\left[  H\right]
}\underbrace{hh^{-1}}_{=1}v^{-1}m\\
&  \qquad\left(
\begin{array}
[c]{c}
\text{since }h\in H\subseteq K\left[  H\right]  \text{ and thus }h\text{ can
be moved}\\
\text{past the tensor sign}
\end{array}
\right) \\
&  =v\otimes_{K\left[  H\right]  }v^{-1}m=Z_{v}\left(  m\right)
\end{align*}
(since the definition of $Z_{v}$ yields $Z_{v}\left(  m\right)  =v\otimes
_{K\left[  H\right]  }v^{-1}m$). In other words, $Z_{u}=Z_{v}$. This proves \eqref{darij1.pf.t2.3}.]
Now, for any left coset $\alpha\in G/H$, we can define a $K$-linear map
$Z_{\alpha}:M\rightarrow K\left[  G\right]  \otimes_{K\left[  H\right]  }M$ by
picking any representative $u\in\alpha$ and setting $Z_{\alpha}=Z_{u}$. This
is well-defined, since $Z_{u}$ does not depend on the choice of representative
$u$ (in fact, \eqref{darij1.pf.t2.3} shows that any two representatives $u$
and $v$ yield the same map $Z_{u}=Z_{v}$).
Recall that the set $G/H$ of left cosets is a left $G$-set, with the action
being defined by $u\cdot\left(  gH\right)  =\left(  ug\right)  H$ for all
$u\in G$ and $g\in G$. Next, we claim that any $g\in G$, $\alpha\in G/H$ and
$m\in M$ satisfy
\begin{equation}
Z_{\alpha}\left(  gm\right)  =g\cdot Z_{g^{-1}\alpha}\left(  m\right)  .
\label{darij1.pf.t2.5}
\tag{2}
\end{equation}
[Proof of \eqref{darij1.pf.t2.5}: Let $g\in G$, $\alpha\in G/H$ and $m\in
M$. Fix any representative $u$ of the left coset $\alpha\in G/H$. Thus,
$u\in\alpha$, so that $Z_{\alpha}=Z_{u}$ (by the definition of $Z_{\alpha}$).
Also, from $u\in\alpha$, we obtain $\alpha=uH$ (since $\alpha$ is a left coset
of $H$) and thus $g^{-1}\alpha=g^{-1}uH$, so that $g^{-1}u\in g^{-1}\alpha$.
Hence, the definition of $Z_{g^{-1}\alpha}$ yields $Z_{g^{-1}\alpha}
=Z_{g^{-1}u}$.
The definition of $Z_{g^{-1}u}$ yields
\begin{equation}
Z_{g^{-1}u}\left(  m\right)  =g^{-1}u\otimes_{K\left[  H\right]
}\underbrace{\left(  g^{-1}u\right)  ^{-1}}_{=u^{-1}g}m=g^{-1}u\otimes
_{K\left[  H\right]  }u^{-1}gm.
\end{equation}
Hence,
\begin{align*}
g\cdot\underbrace{Z_{g^{-1}u}\left(  m\right)  }_{=g^{-1}u\otimes_{K\left[
H\right]  }u^{-1}gm}  &  =g\cdot\left(  g^{-1}u\otimes_{K\left[  H\right]
}u^{-1}gm\right)  =\underbrace{\left(  gg^{-1}\right)  }_{=1}u\otimes
_{K\left[  H\right]  }u^{-1}gm\\
&  =u\otimes_{K\left[  H\right]  }u^{-1}gm.
\end{align*}
Comparing this with
\begin{equation}
Z_{u}\left(  gm\right)  =u\otimes_{K\left[  H\right]  }u^{-1}gm\qquad\left(
\text{by the definition of }Z_{u}\right)  ,
\end{equation}
we obtain $Z_{u}\left(  gm\right)  =g\cdot Z_{g^{-1}u}\left(  m\right)  $. In
view of $Z_{\alpha}=Z_{u}$ and $Z_{g^{-1}\alpha}=Z_{g^{-1}u}$, this rewrites
as $Z_{\alpha}\left(  gm\right)  =g\cdot Z_{g^{-1}\alpha}\left(  m\right)  $.
This proves \eqref{darij1.pf.t2.5}.]
Now, define a $K$-linear map
\begin{equation}
P:M\rightarrow K\left[  G\right]  \otimes_{K\left[  H\right]  }M
\end{equation}
by
\begin{equation}
P=\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in G/H}Z_{\alpha}.
\label{darij1.def.T.2}
\tag{3}
\end{equation}
(This is well-defined, since $\left[  G:H\right]  $ is invertible in $K$.)
I claim that this map $P$ is a (left) $K\left[  G\right]  $-module homomorphism.
[Proof: Let $g\in G$ and $m\in M$. Recall that $G/H$ is a left $G$-set;
thus, the map $G/H\rightarrow G/H,\ \alpha\mapsto g^{-1}\alpha$ is a bijection.
Applying both sides of \eqref{darij1.def.T.2} to $m$, we obtain
\begin{equation}
P\left(  m\right)  =\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in
G/H}Z_{\alpha}\left(  m\right)  .
\label{darij1.pf.P-equiv.1}
\tag{4}
\end{equation}
But applying both sides of \eqref{darij1.def.T.2} to $gm$, we obtain
\begin{align*}
P\left(  gm\right)   &  =\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in
G/H}\underbrace{Z_{\alpha}\left(  gm\right)  }_{\substack{=g\cdot
Z_{g^{-1}\alpha}\left(  m\right)  \\\text{(by \eqref{darij1.pf.t2.5})}
}}=\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in G/H}g\cdot Z_{g^{-1}\alpha
}\left(  m\right) \\
&  =\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in G/H}g\cdot Z_{\alpha
}\left(  m\right) \\
&  \qquad\left(
\begin{array}
[c]{c}
\text{here, we have substituted }\alpha\text{ for }g^{-1}\alpha\text{ in the
sum,}\\
\text{since the map }G/H\rightarrow G/H,\ \alpha\mapsto g^{-1}\alpha\text{ is
a bijection}
\end{array}
\right) \\
&  =g\cdot\underbrace{\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in
G/H}Z_{\alpha}\left(  m\right)  }_{\substack{=P\left(  m\right)  \\\text{(by
\eqref{darij1.pf.P-equiv.1})}}}=g\cdot P\left(  m\right)  .
\end{align*}
Now, forget that we fixed $g$ and $m$. We thus have shown that $P\left(
gm\right)  =g\cdot P\left(  m\right)  $ for all $g\in G$ and $m\in M$. In
other words, the map $P$ is $G$-equivariant. Since $P$ is also $K$-linear, we
thus conclude that $P$ is a $K\left[  G\right]  $-module homomorphism. Qed.]
On the other hand, we define a $K$-linear map
\begin{equation}
Q:K\left[  G\right]  \otimes_{K\left[  H\right]  }M\rightarrow M
\end{equation}
by
\begin{equation}
Q\left(  u\otimes_{K\left[  H\right]  }m\right)  =um\qquad\text{for all }u\in
K\left[  G\right]  \text{ and }m\in M.
\end{equation}
This is clearly well-defined.
We claim that this map $Q$ is a (left) $K\left[  G\right]  $-module homomorphism.
[Proof: Let $v\in K\left[  G\right]  $ and $r\in K\left[  G\right]
\otimes_{K\left[  H\right]  }M$. We shall prove that $Q\left(  vr\right)
=v\cdot Q\left(  r\right)  $.
Indeed, both sides of this equality are $K$-linear in $r$; hence, we can WLOG
assume that $r$ is a pure tensor (since the $K$-module $K\left[  G\right]
\otimes_{K\left[  H\right]  }M$ is spanned by pure tensors). Assume this, and
write $r$ as $u\otimes_{K\left[  H\right]  }m$ for some $u\in K\left[
G\right]  $ and $m\in M$. Thus, $r=u\otimes_{K\left[  H\right]  }m$, so that
$vr=v\left(  u\otimes_{K\left[  H\right]  }m\right)  =vu\otimes_{K\left[
H\right]  }m$. Hence,
\begin{equation}
Q\left(  vr\right)  =Q\left(  vu\otimes_{K\left[  H\right]  }m\right)
=vum\qquad\left(  \text{by the definition of }Q\right)  .
\end{equation}
Comparing this with
\begin{equation}
v\cdot Q\left(  \underbrace{r}_{=u\otimes_{K\left[  H\right]  }m}\right)
=v\cdot\underbrace{Q\left(  u\otimes_{K\left[  H\right]  }m\right)
}_{\substack{=um\\\text{(by the definition of }Q\text{)}}}=vum,
\end{equation}
we obtain $Q\left(  vr\right)  =v\cdot Q\left(  r\right)  $.
Now, forget that we fixed $v$ and $r$. We thus have shown that $Q\left(
vr\right)  =v\cdot Q\left(  r\right)  $ for all $v\in K\left[  G\right]  $ and
$r\in K\left[  G\right]  \otimes_{K\left[  H\right]  }M$. In other words, the
map $Q$ is a (left) $K\left[  G\right]  $-module homomorphism (since $Q$ is
$K$-linear). Qed.]
Furthermore, we claim that
\begin{equation}
Q\circ Z_{\alpha}=\operatorname*{id}\qquad\text{for each }\alpha\in G/H.
\label{darij1.pf.t2.9}
\tag{5}
\end{equation}
[Proof of \eqref{darij1.pf.t2.9}: Let $\alpha\in G/H$. Fix any
representative $u$ of the left coset $\alpha\in G/H$. Thus, $u\in\alpha$, so
that $Z_{\alpha}=Z_{u}$ (by the definition of $Z_{\alpha}$). Now, for each
$m\in M$, we have
\begin{align*}
Z_{\alpha}\left(  m\right)   &  =Z_{u}\left(  m\right)  \qquad\left(
\text{since }Z_{\alpha}=Z_{u}\right) \\
&  =u\otimes_{K\left[  H\right]  }u^{-1}m\qquad\left(  \text{by the definition
of }Z_{u}\right)
\end{align*}
and thus
\begin{align*}
\left(  Q\circ Z_{\alpha}\right)  \left(  m\right)   &  =Q\left(
\underbrace{Z_{\alpha}\left(  m\right)  }_{=u\otimes_{K\left[  H\right]
}u^{-1}m}\right)  =Q\left(  u\otimes_{K\left[  H\right]  }u^{-1}m\right) \\
&  =\underbrace{uu^{-1}}_{=1}m\qquad\left(  \text{by the definition of
}Q\right) \\
&  =m=\operatorname*{id}\left(  m\right)  .
\end{align*}
In other words, $Q\circ Z_{\alpha}=\operatorname*{id}$. This proves \eqref{darij1.pf.t2.9}.]
Now, we can easily show that
\begin{equation}
Q\circ P=\operatorname*{id}.
\label{darij1.pf.t2.10}
\tag{6}
\end{equation}
[Proof of \eqref{darij1.pf.t2.10}: We have $P=\dfrac{1}{\left[  G:H\right]
}\sum_{\alpha\in G/H}Z_{\alpha}$ (by \eqref{darij1.pf.t2.9}) and thus
\begin{align*}
Q\circ P  &  =Q\circ\left(  \dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in
G/H}Z_{\alpha}\right) \\
&  =\dfrac{1}{\left[  G:H\right]  }\sum_{\alpha\in G/H}\underbrace{Q\circ
Z_{\alpha}}_{\substack{=\operatorname*{id}\\\text{(by \eqref{darij1.pf.t2.9})}
}}\qquad\left(  \text{since the map }Q\text{ is }K\text{-linear}\right) \\
&  =\dfrac{1}{\left[  G:H\right]  }\underbrace{\sum_{\alpha\in G/H}
\operatorname*{id}}_{=\left\vert G/H\right\vert \cdot\operatorname*{id}
}=\underbrace{\dfrac{1}{\left[  G:H\right]  }\left\vert G/H\right\vert
}_{\substack{=1\\\text{(since }\left[  G:H\right]  =\left\vert G/H\right\vert
\text{)}}}\cdot\operatorname*{id}=\operatorname*{id}.
\end{align*}
This proves \eqref{darij1.pf.t2.10}.]
Altogether, we thus have found two $K\left[  G\right]  $-module homomorphisms
$P:M\rightarrow K\left[  G\right]  \otimes_{K\left[  H\right]  }M$ and
$Q:K\left[  G\right]  \otimes_{K\left[  H\right]  }M\rightarrow M$ that
satisfy $Q\circ P=\operatorname*{id}$. Hence, Lemma 3 (applied to $A=K\left[
G\right]  $, $U=M$ and $V=K\left[  G\right]  \otimes_{K\left[  H\right]  }M$)
shows that $M$ appears as a direct addend in the $K\left[  G\right]  $-module
$K\left[  G\right]  \otimes_{K\left[  H\right]  }M$. In other words, $M$
appears as a direct addend in the $K\left[  G\right]  $-module
$\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$
(since $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}
^{G}M=K\left[  G\right]  \otimes_{K\left[  H\right]  }M$). This proves Theorem
2. $\blacksquare$
Proof of Theorem 1. $\Longrightarrow:$ Assume that $M$ is projective as a
$K\left[  G\right]  $-module. Thus, $M$ is a direct addend of a free $K\left[
G\right]  $-module. In other words, $M$ is a direct addend of $\left(
K\left[  G\right]  \right)  ^{\left(  I\right)  }$ for some set $I$. Consider
this $I$.
So we know that $M$ is a direct addend of $\left(  K\left[  G\right]  \right)
^{\left(  I\right)  }$. In other words, there exists a $K\left[  G\right]
$-module $N$ such that $M\oplus N\cong\left(  K\left[  G\right]  \right)
^{\left(  I\right)  }$. Consider this $N$.
The (left) $K\left[  H\right]  $-module $\operatorname{Res}_{H}
^{G}\left(  K\left[  G\right]  \right)  $ is precisely the $K\left[  H\right]
$-module $K\left[  G\right]  $ on which $K\left[  H\right]  $ acts by left
multiplication. This $K\left[  H\right]  $-module $K\left[  G\right]  $ is
free (indeed, if we pick any system $R$ of representatives of the right
$H$-cosets in $G$, so that $G=\bigsqcup\limits_{r\in R}Hr$, then $K\left[
G\right]  $ is a free left $K\left[  H\right]  $-module with basis $\left(
r\right)  _{r\in R}$). In other words, there exists a set $R$ such that
$K\left[  G\right]  \cong\left(  K\left[  H\right]  \right)  ^{\left(
R\right)  }$ as a $K\left[  H\right]  $-module. Consider this $R$.
Now, $K\left[  G\right]  \cong\left(  K\left[  H\right]  \right)  ^{\left(
R\right)  }$ as a $K\left[  H\right]  $-module. But restriction of modules
respects direct sums; thus,
\begin{align*}
\left(  \operatorname{Res}_{H}^{G}M\right)  \oplus\left(
\operatorname{Res}_{H}^{G}N\right)   &  \cong\operatorname{Res}_{H}^{G}
\underbrace{\left(  M\oplus N\right)  }_{\cong\left(
K\left[  G\right]  \right)  ^{\left(  I\right)  }}\cong
\operatorname{Res}_{H}^{G}\left(  \left(  K\left[  G\right]  \right)  ^{\left(
I\right)  }\right)  \\
&  \cong\left(  \underbrace{\operatorname{Res}_{H}^{G}\left(
K\left[  G\right]  \right)  }_{=K\left[  G\right]  \cong\left(  K\left[
H\right]  \right)  ^{\left(  R\right)  }}\right)  ^{\left(  I\right)  }
\cong\left(  \left(  K\left[  H\right]  \right)  ^{\left(  R\right)  }\right)
^{\left(  I\right)  }\\
&  \cong\left(  K\left[  H\right]  \right)  ^{\left(  R\times I\right)  }.
\end{align*}
Hence, $\operatorname{Res}_{H}^{G}M$ is a direct addend of a free
$K\left[  H\right]  $-module (namely, of $\left(  K\left[  H\right]  \right)
^{\left(  R\times I\right)  }$), and thus is projective as a $K\left[
H\right]  $-module. This proves the "$\Longrightarrow$" direction of Theorem 1.
$\Longleftarrow:$ Assume that the restriction
$\operatorname{Res}_{H}^{G}M$ is projective as a
$K\left[  H\right]  $-module. In other
words, $\operatorname{Res}_{H}^{G}M$ is a direct addend of a free
$K\left[  H\right]  $-module. In other words, there exist a set $I$ and a
$K\left[  H\right]  $-module $N$ such that $\left( 
\operatorname{Res}_{H}^{G}M\right)  \oplus N\cong\left(  K\left[  H\right]  \right)
^{\left(  I\right)  }$. Consider these $I$ and $N$.
The definition of induced modules yields $\operatorname{Ind}_{H}
^{G}\left(  K\left[  H\right]  \right)  =K\left[  G\right]  \otimes_{K\left[
H\right]  }K\left[  H\right]  \cong K\left[  G\right]  $. But induction of
modules respects direct sums; thus,
\begin{align*}
&  \left(  \operatorname{Ind}_{H}^{G}
\operatorname{Res}_{H}^{G}M\right)  \oplus\left(  \operatorname{Ind}_{H}
^{G}N\right)  \\
&  \cong\operatorname{Ind}_{H}^{G}\underbrace{\left(  \left(
\operatorname{Res}_{H}^{G}M\right)  \oplus N\right)  }_{\cong\left(
K\left[  H\right]  \right)  ^{\left(  I\right)  }}\cong
\operatorname{Ind}_{H}^{G}\left(  \left(  K\left[  H\right]  \right)  ^{\left(
I\right)  }\right)  \cong\left(
\underbrace{\operatorname{Ind}_{H}^{G}\left(  K\left[  H\right]  \right)
}_{\cong K\left[
G\right]  }\right)  ^{\left(  I\right)  }\\
&  \cong\left(  K\left[  G\right]  \right)  ^{\left(  I\right)  }.
\end{align*}
Hence, $\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}
^{G}M$ appears as a direct addend in the $K\left[  G\right]  $-module $\left(
K\left[  G\right]  \right)  ^{\left(  I\right)  }$. Since $M$, in turn,
appears as a direct addend in the $K\left[  G\right]  $-module
$\operatorname{Ind}_{H}^{G}\operatorname{Res}_{H}^{G}M$
(by Theorem 2), we thus conclude that $M$ appears as a direct addend in the
$K\left[  G\right]  $-module $\left(  K\left[  G\right]  \right)  ^{\left(
I\right)  }$ (since the binary relation "appears as a direct addend in" is
transitive). Thus, $M$ appears as a direct addend in a free $K\left[
G\right]  $-module (since $\left(  K\left[  G\right]  \right)  ^{\left(
I\right)  }$ is clearly a free $K\left[  G\right]  $-module). In other words,
$M$ is projective as a $K\left[  G\right]  $-module. This proves the
"$\Longleftarrow$" direction of Theorem 1. $\blacksquare$
