The complex case then follows from (a).

Continuity then finishes off the argument.

Theorem 3 may be interpreted as saying that $A=B$, but it must then be remembered that ......

Then $G$ has 10 normal subgroups and as many non-normal ones.

If $p=0$ then there are an additional $m$ arcs. [Note the article an.]

If $y$ is a solution, then $ay$ also solves (3) for all $a$ in $B$.

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