Groups of Order 30/Mistake

Source Work

1996: John F. Humphreys: A Course in Group Theory:

Chapter $13$: Direct products:

Theorem $13.8$

Mistake

To show that $x$ normalises $K$, note that

$x x^5 x^{-1} = x$, and $x y x^{-1} = x y^{-1} x^{-1} = y x^{11} x^{-1} = y x^{10} \in N$.

Correction

There are two mistakes here:

$(1): \quad x x^5 x^{-1} = x$ makes no sense in this context. Here we have $x^5 = w$, which we are trying to show is the conjugate of $x$.

So it is apparent that what is meant is more likely something along the lines of:

$x w x^{-1} = x x^5 x^{-1} = x^5 = w$

$(2): \quad$ That last $N$ looks as though it ought to be $K$, as there is no $N$ defined in this section.

Sources

• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $13$: Direct products: Theorem $13.8$