2

The question is from Herstein. Until now, semi direct product has not been covered in the material. I am solving these type of questions using the procedure given by Herstein in previous sections. I was able to solve it for order 55 but for order 203, I am facing an issue. The procedure goes like this ::

Required Order = 203 = 29 * 7

Consider a cyclic group G of order 29.

G = $<a>$ such that $a^{29} = e$

And consider an automorphic mapping $\phi$: $a^i \to a^{6i}$. Ideally order of $\phi$ should be 7 here. i.e. $\phi ^7 = I$.

But In this case, $\phi ^7 = a^{279936i} = a^{-i}$.

If $\phi ^7=I$ was true in this case, then next steps would be to consider a group containing all formal elements $x^ia^j$. This group is the required group asked for in the question which is of order 203.

where i = 0,1,2,3,4,5,6. and j = 0,1,2, ...... ,28.

And symbol x is subject to few conditions which are ::

$x^7=e$

$x^{-1}a^ix = a^{6i}$

$x^ia^j = x^ka^l$ iff i $\equiv$ k (mod 7) and j $\equiv$ l (mod 29).

Please let me know where I am going wrong, why $\phi^7$ isnot equal to I ?

latus_rectum
  • 139
  • 1
    It might be easier to look for the general case of order $pq$, see for example here. – Dietrich Burde Nov 16 '20 at 10:07
  • I've no idea why you chose $a\mapsto a^6$. I think $a\mapsto a^5$ might work. – ancient mathematician Nov 16 '20 at 10:54
  • @ancientmathematician, It's based on the factorization of required group. Since 203 = 29*7, therefore the order of the automorphic mapping should be 7. And for $a^5$, $\phi^7$ will be $a^{78125i}$, again it's not equal to $a^i$ – latus_rectum Nov 16 '20 at 12:09
  • 1
    Sorry, I made a slip, $a\mapsto a^5$ taken seven times inverts $a$ [as you say] so we need to square it. Try $a\mapsto a^{-4}$. – ancient mathematician Nov 16 '20 at 12:42
  • $a^{-4}$ will not work. But now that you have given me this hint of mapping it to a negative power. I was doing $a^i \to a^{6i}$. Instead if I do, $a^i \to a^{-6i}$, then $\phi^7$ = $a^{-279936i}$ = $a^{-279937i}a^i$ = $a^{-(299653)i}a^i$ = $e^{-1}a^i$ = I. Thus the order of $\phi$ is 7. Thanks. – latus_rectum Nov 16 '20 at 13:13
  • @quasi, how did you came up with list of $k$ ? – latus_rectum Nov 16 '20 at 14:16
  • 1
    Correcting my earlier (deleted) comment, $a^k$ works for $k\in{7,16,20,23,24,25}$ or equivalently $k\in{7,-13,-9,-6,-5,-4}$. We want $k$ to be of order $7$ mod $29$, so if we find a primitive root $r$ (e.g., $r=2$), then the set of elements of order $7$ is the set ${r^{4j};\text{mod};29:1\le j\le 6}$. – quasi Nov 16 '20 at 14:55

0 Answers0