I have the following definition of perfect secrecy (please assume that the probabilistic version is not available):
If we consider the eavesdropping game given by:
$$\begin{array}{|r | r|} \hline Alice & Eve \\ \hline & (m_0,m_1) = E.m()\\ k \leftarrow Gen() & \ \\ b \leftarrow fair\_coin() & \\ c \leftarrow Enc_k (m_b) & \\ & r = E.r(c)\\ \hline \end{array}$$
one says that a symmetric encription system is perfectly secret if there is no attack $E$ which wins the game with probability strictly greater than $1/2$, where winning the game means that at the end of the process $b = r$.
With this definition I have to prove that the $(G,l)$-one time pad (that is one-time pad over groups) given by $Enc_k(m) = m \circ k$ and $Dec_k(c) = c \circ k^{-1}$ is perfectly secret. Here one defines for $m = (m_1,\ldots,m_l)$ and $k = (k_1,\ldots,k_l)$, $m \circ k = (m_1 \circ k_1,\ldots,m_l \circ k_l)$ and $k^{-1} = (k_1^{-1},\ldots,k_l^{-1})$.
My approach
$k$ is uniformly distributed by assumption and since $m_b$ should be independent of $k$ then $m_b \circ k$ is uniformly distributed. But no attack can distinguish in a uniformly distributed sequence with probability greater than 2.
Is my approach correct?