## Every group of order 203 with a normal subgroup of order 7 is abelian

Let be a group of order 203. Prove that if is a normal subgroup of order 7 then . Prove that is abelian.

Note that . Now , so that . Moreover, since has prime order it is cyclic, hence abelian, and so , and thus (by Lagrange) divides 29. Now by Proposition 17. Since 29 and 6 are relatively prime, we have , hence . Thus .

Now is maximal. If , we have cyclic since (by Lagrange) is prime. Then is abelian, so that ; a contradiction. Thus , and we have abelian.

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## Comments

In the second part we said :” Now H =< G is maximal. If Z(G)= H , we have G/Z(G) cyclic since |G/Z(G)| (by Lagrange) is prime. Then G is abelian, so that Z(G)=G ; a contradiction."

Then you said:" Thus ,Z(G)=G and we have abelian."

I am confused why you said it was a contradiction, then you concluded again Z(G)=G. Is there just a mistyped!!!

The first part shows that is either or . By assuming that , we get (via the theorem) that . This is a contradiction of the assumption that . Thus, .

can you explain why G/C(H) has to divide 29. this seems to be the case in the other 2 problems and I can’t recall why. thanks

This is because for finite groups , .

Sorry, but I’m confused as well about this. Why can we say divides 29, shouldn’t we say ? A priori, you do not know if the size of is 1, 7, 29, or 203, so for all we know $|G/C_{G}(H)| = 203$ which does NOT divide 29.

I think a less hand-wavy path would have involved considering the size of the stabilizer of in and showing that the only value that works is 203.

Since is abelian (being of prime order) it is contained in its own centralizer.

Cool! Thanks. That fact totally slipped past me.

I was confused by the extra-explanation compared with the previous problem. And I noticed that it was because of the ask in the comments. But actually we only need to know that divides 203.