Orders of p-groups, normal subgroups

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April 11th 2010, 01:37 PM
kimberu

Orders of p-groups, normal subgroups

If H is a p-group show every element can be multiplied by some power to equal p.
April 11th 2010, 02:22 PM
FancyMouse

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 $x^{p^{r-1}}$ ?
b. Further hint: prove that $x\in N$ and $x\in L$ , 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 $A\cap B=\{e\}$ and everything in A commutes with everything in B)
April 11th 2010, 02:40 PM
maddas

(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?
April 11th 2010, 03:25 PM
kimberu

Quote:

Originally Posted by FancyMouse

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 $x^{p^{r-1}}$ ?
b. Further hint: prove that $x\in N$ and $x\in L$ , 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 $A\cap B=\{e\}$ and everything in A commutes with everything in B)

Thank you! I understand (c), and I think for (b), after showing x is in N and L, this implies that x = 1 = $aba^{-1}b^{-1} = b(aa^{-1})b^{-1}$ 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 $x^{p^{r-1}}$ ?
April 11th 2010, 05:06 PM
aliceinwonderland

Quote:

Originally Posted by kimberu

Thank you! I understand (c), and I think for (b), after showing x is in N and L, this implies that x = 1 = $aba^{-1}b^{-1} = b(aa^{-1})b^{-1}$ 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 $x^{p^{r-1}}$ ?

Well, that is not correct.

$aba^{-1} \in L$ and $(aba^{-1})b^{-1} \in L$ .
$ba^{-1}b^{-1} \in N$ and $a(ba^{-1}b^{-1}) \in N$ .Then, $aba^{-1}b^{-1} \in N \cap L = \{e\}$ .
Thus $ab=ba$ for all $a \in N, b \in L$ .
April 11th 2010, 05:21 PM
kimberu

Quote:

Originally Posted by aliceinwonderland

Well, that is not correct.

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

Sorry, I'm still unclear -- Why does $aba^{-1}b^{-1} \in N \cap L = \{e\}$ imply that ab =ba, if not for the reason I stated before?
April 11th 2010, 05:26 PM
aliceinwonderland

Quote:

Originally Posted by kimberu

Sorry, I'm still unclear -- Why does $aba^{-1}b^{-1} \in N \cap L = \{e\}$ imply that ab =ba, if not for the reason I stated before?

$aba^{-1}b^{-1} = 1 \leftrightarrow aba^{-1}= b \leftrightarrow ab=ba$ .