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Math Help - Orders of p-groups, normal subgroups

  1. #1
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    Orders of p-groups, normal subgroups

    If H is a p-group show every element can be multiplied by some power to equal p.
    Last edited by kimberu; April 12th 2010 at 08:05 PM.
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    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)
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  3. #3
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    (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?
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  4. #4
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    Quote Originally Posted by FancyMouse View Post
    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}}?
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  5. #5
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    Quote Originally Posted by kimberu View Post
    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.
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  6. #6
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    Quote Originally Posted by aliceinwonderland View Post
    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?
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  7. #7
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    Quote Originally Posted by kimberu View Post
    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 .
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