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Understanding semicolons in Bash functions

I know I can write a function at a Bash command line like so:

```
$ test() { echo 'test' ; }
```

I also found that I can write the `}` on a separate line, and omit `;`:

```
$ test () { echo 'test'
> }
```

But I *cannot* omit `;` when inputting on the same line:

```
$ test () { echo 'test' }
> 
> 
```

Here, the interface for inputting a multi-line function doesn't exit until I either cancel it manually, input a `;` (resulting in an immediate syntax error), or input a `}` (resulting in an extra `}` in the function (now the `}` becomes an argument to `echo` instead).

On the other hand, if I want the process to be backgrounded, I need not and indeed *cannot* use `;`:

```
$ test () { echo 'test' & }
```

creates the function immediately which works as expected, but 

```
$ test () { echo 'test' & ;
```

is an immediate syntax error (giving no chance to input the balancing `}`), and is still a syntax error with a `}` on the same line. I tried swapping the `&` and `;` but with no meaningful effect.

**What does `;` actually do in a Bash function?**

Consequently: Why isn't `}` on the same line recognized as a closing brace for the function without either `;` or `&` at the end of the command? Why *is* it recognized as such on a *separate* line? Why do I need `;` for an ordinary command invocation, but not for a backgrounded one?