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Anyone have an idea on how to do this problem? | Anyone have an idea on how to do this problem? | ||
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| + | I got the idea from Uli in office hours, but I am struggling with the details. Here is the idea: U(25) has order 20. Start this problem VERY similarly to the example in the notes from 2/19/09 where Uli has a group G with order 16 and he uses a theorem from those notes to come up G= U(65)= order 16 | ||
| + | So G=Z/16Z or Z/2Z x Z/8/Z or..... | ||
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| + | Okay, follow that, but in this case, U(25)= Z/20Z or Z/2Z X Z/10Z or Z/4Z x Z/5Z or Z/2Z x Z/2Z x Z/5Z. Then we realize that <2> is a cyclic generator of U(25) because <2> has 20 elements. Okay, here is where I get lost in the details, but we want to find out which Z modulars do work for U(25). We find that it is Z/20Z which is also equal to Z/4Z X Z/5Z. The other Z modular choices we had listed do not work. Okay, now we can say (somehow) that | ||
| + | Aut(U(25))=Aut(Z/20Z). And From Thm 6.5 we know Aut(Z/20Z) = U(20). | ||
| + | Good, now we can say U(20)=U(4x5) because 4 and 5 are relatively prime to each other(Thm.8.3). | ||
| + | So U(20) = U(4)x U(5). We can rewrite U(5) as Zsubscript4 and U(4) as Zsubscript2. | ||
| + | Thus, Aut(U(25)) is now in the form Zsubscript4 x Zsubscript2. | ||
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| + | That is the rough idea. I didn't follow along well enough to what he said to have all of the details, but I hope that that helps. | ||
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| + | ---- | ||
| + | Oh wow... I should have thought about that much earlier... I think I actually know the details that you don't remember. Thanks a lot. | ||
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| + | [[Category:MA453Spring2009Walther]] | ||
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| + | I still am not really sure how this problem works out but this explain does help somewhat. I am confused on the same part as you that Aut(U(25)= Aut(z/20z). That's where I got stuck too. Thanks though for the rest of the info. | ||
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| + | --[[User:Johns121|Johns121]] 19:19, 26 February 2009 (UTC) | ||
Latest revision as of 14:19, 26 February 2009
Anyone have an idea on how to do this problem?
I got the idea from Uli in office hours, but I am struggling with the details. Here is the idea: U(25) has order 20. Start this problem VERY similarly to the example in the notes from 2/19/09 where Uli has a group G with order 16 and he uses a theorem from those notes to come up G= U(65)= order 16 So G=Z/16Z or Z/2Z x Z/8/Z or.....
Okay, follow that, but in this case, U(25)= Z/20Z or Z/2Z X Z/10Z or Z/4Z x Z/5Z or Z/2Z x Z/2Z x Z/5Z. Then we realize that <2> is a cyclic generator of U(25) because <2> has 20 elements. Okay, here is where I get lost in the details, but we want to find out which Z modulars do work for U(25). We find that it is Z/20Z which is also equal to Z/4Z X Z/5Z. The other Z modular choices we had listed do not work. Okay, now we can say (somehow) that Aut(U(25))=Aut(Z/20Z). And From Thm 6.5 we know Aut(Z/20Z) = U(20). Good, now we can say U(20)=U(4x5) because 4 and 5 are relatively prime to each other(Thm.8.3). So U(20) = U(4)x U(5). We can rewrite U(5) as Zsubscript4 and U(4) as Zsubscript2. Thus, Aut(U(25)) is now in the form Zsubscript4 x Zsubscript2.
That is the rough idea. I didn't follow along well enough to what he said to have all of the details, but I hope that that helps.
Oh wow... I should have thought about that much earlier... I think I actually know the details that you don't remember. Thanks a lot.
I still am not really sure how this problem works out but this explain does help somewhat. I am confused on the same part as you that Aut(U(25)= Aut(z/20z). That's where I got stuck too. Thanks though for the rest of the info.
--Johns121 19:19, 26 February 2009 (UTC)
