If H is a p-group show every element can be multiplied by some power to equal p.

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- Apr 11th 2010, 01:37 PMkimberuOrders of p-groups, normal subgroups
If H is a p-group show every element can be multiplied by some power to equal p.

- Apr 11th 2010, 02:22 PMFancyMouse
a. No need to bring in Cauchy. For any x in G, if x is not the identity, then x has an order p^r for some r>0. What about $\displaystyle x^{p^{r-1}}$?

b. Further hint: prove that $\displaystyle x\in N$ and $\displaystyle x\in L$, using the fact that N,L are normal in G

c. Suppose A is the subgroup of N of order 2, B is the subgroup of L of order 3. We know that everything in N and L commutes. In particular, everything in A commutes with everything in B. But A and B are cyclic. What is the subgroup AB look like? (Notice that the "subgroup" AB makes sense because $\displaystyle A\cap B=\{e\}$ and everything in A commutes with everything in B) - Apr 11th 2010, 02:40 PMmaddas
(b) is so

*intuitive*to me, but the proof your supposed to find is so*opaque*. Can't you use that G is the semidirect product of N and L somehow? - Apr 11th 2010, 03:25 PMkimberu
Thank you! I understand (c), and I think for (b), after showing x is in N and L, this implies that x = 1 = $\displaystyle aba^{-1}b^{-1} = b(aa^{-1})b^{-1}$ which implies ab = ba? (Or am I completely wrong?

But I'm still lost on part (a), could you possibly elaborate further? What is the significance of $\displaystyle x^{p^{r-1}}$? - Apr 11th 2010, 05:06 PMaliceinwonderland
Well, that is not correct.

$\displaystyle aba^{-1} \in L$ and $\displaystyle (aba^{-1})b^{-1} \in L$.

$\displaystyle ba^{-1}b^{-1} \in N$ and $\displaystyle a(ba^{-1}b^{-1}) \in N$.Then, $\displaystyle aba^{-1}b^{-1} \in N \cap L = \{e\}$.

Thus $\displaystyle ab=ba$ for all $\displaystyle a \in N, b \in L$. - Apr 11th 2010, 05:21 PMkimberu
- Apr 11th 2010, 05:26 PMaliceinwonderland