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This is an example from Hungerford's textbook ''Algebra''.

Let G and H be multiplicative groups and $f:G \rightarrow H$ a homomorphism of groups. Let R be a ring and define a map on the group rings $\bar{f}:R(G) \rightarrow R(H)$ by: $$\bar{f} \left( \sum_{i=1}^{n}r_i g_i \right) =\sum_{i=1}^{n}r_i f(g_i) $$

Then, $\bar{f}$ is a homomorphism of rings.

I know that we have to show that given $a, b \in R(G)$, $\bar{f}(a + b) = \bar{f}(a) + \bar{f}(b)$ and $\bar{f}(a b) = \bar{f}(a) \bar{f}(b)$.

Given $a = \sum_{i=1}^{n}r_ig_i$ and $b=\sum_{i=1}^{n}s_ig_i$, I showed that $\bar{f}(a + b) = \bar{f}(a) + \bar{f}(b)$.

However, I got stuck trying to show that $\bar{f}(a b) = \bar{f}(a) \bar{f}(b)$.

This is how I did it so far:

$\bar{f}\left(\sum_{i=1}^{n}r_ig_i \cdot \sum_{i=1}^{n}s_ig_i \right) $ = $\bar{f}\left(\sum_{i=1}^{n} \sum_{i=1}^{n}(r_is_i)(g_ig_i) \right) $

= $\sum_{i=1}^{n} \sum_{i=1}^{n}(r_is_i)f(g_ig_i)$

= $\sum_{i=1}^{n} \sum_{i=1}^{n}(r_is_i)f(g_i)f(g_i)$

I'm not sure it is the correct approach. And if it is, I don't know to finish it, given that neither the groups nor the rings are required to be commutative.

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    You should use different indices, as $(r_is_j)(g_ig_j)$ instead of $(r_is_i)(g_ig_i)$ – 2017-02-10
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    And you're basically done. The last thing you wrote is $$(\sum\limits_i r_if(g_i))(\sum\limits_j s_j f(g_j))$$ which is $\overline{f}(a) \overline{f}(b)$. – 2017-02-10

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