MHB Conjugating Subgroups: Proving |X_K| Divides |K|

[Kiwi1]

I have answered all parts of the following question except for the very last sentence:
'Conclude that the number of elements in X_K is a divisor of |K|.'

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MY THOUGHTS
Presumably I must argue that ord(K*) divides ord(K).

Clearly Ord(K*) =< ord (K).
Also I can show that for any element Na in K*: ord (Na) divides ord (K)
But this is not sufficient.

N and K are both subgroups of G. But I know nothing of any relationship between N and K other than that they have the identity element in common.

Any ideas?

 

[Euge]

Hi Kiwi,

Since $X_K$ is in one-to-one correspondence with $K^*$, it has just as many elements as $K^*$. How many elements are in $K^*$? (Use Problem 8).

 

[Kiwi1]

I don't see how to use question 8 (yet). But does this work?

Define a function g from K to K* such that g(a)=Na
this is an onto homomorphism with Kernel (N intersection K), so by the isomorphism theorem:
K/(N intersection K) is isomorphic to K*

But ord (K/(N intersection K)) divides ord (K)

so ord(K*) divides ord (K)

therefore ord (X_K) divides ord(K)

Edit: that does not work because the isomorphism theorem is not introduced for another 2 chapters.