My abstract algebra book is talking about reducing the calculations involved in determining whether a subgroup is normal. It says:(adsbygoogle = window.adsbygoogle || []).push({});

If N is a subgroup of a group G, then N is normal iff for all g in G, gN(g^-1) [the conjugate of N by g] = N.

If one has a set of generators for N, it suffices to check that all conjugates of these generators lie in N to prove that N is a normal subgroup (because the conjugate of a product is the product of the conjugates and the conjugate of the inverse is the inverse of the conjugate).

Similarly, if generators for G are known, then it suffices to check that the conjugates of N by these generators all equal N [these generators for G normalize N].

In particular, if generators for both N and G are known, then this reduces the calculations to a small number of conjugations to check. If N is a FINITE group then it suffices to check that the conjugates of a set of generators for N by a set of generators for G are again elements of N.

My question is: why does it stipulate that N be finite? Wouldn't this work if N has infinite order as well?

Thanks!

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# Method to reduce calculations to det. normal subgroup only works for finite groups?

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