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

Let G be a group of order 203. Prove that if H \leq G is a normal subgroup of order 7 then H \leq Z(G). Prove that G is abelian.


Note that 203 = 7 \cdot 29. Now N_G(H) = G, so that G/C_G(H) \leq \mathsf{Aut}(H). Moreover, since H has prime order it is cyclic, hence abelian, and so H \leq C_G(H), and thus (by Lagrange) |G/C_G(H)| divides 29. Now |\mathsf{Aut}(H)| = 6 by Proposition 17. Since 29 and 6 are relatively prime, we have G/C_G(H) = 1, hence C_G(H) = G. Thus H \leq Z(G).

Now H \leq G is maximal. If Z(G) = H, we have G/Z(G) cyclic since (by Lagrange) |G/Z(G)| is prime. Then G is abelian, so that Z(G) = G; a contradiction. Thus Z(G) = G, and we have G abelian.

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Comments

  • saigonbettas  On December 8, 2010 at 5:55 pm

    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!!!

    • nbloomf  On December 9, 2010 at 9:44 am

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

  • neal  On July 25, 2011 at 7:46 pm

    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

    • nbloomf  On August 5, 2011 at 8:37 am

      This is because for finite groups G, |G/N| = |G|/|N|.

      • Steven  On December 5, 2011 at 8:22 pm

        Sorry, but I’m confused as well about this. Why can we say G/C_{G}(H) divides 29, shouldn’t we say G/C_{G}(H) = (7 \cdot 29)/|C_{G}(H)|? A priori, you do not know if the size of C_{G}(H) 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 H in G and showing that the only value that works is 203.

        • nbloomf  On December 5, 2011 at 8:35 pm

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

          • Steven  On December 5, 2011 at 9:45 pm

            Cool! Thanks. That fact totally slipped past me.

  • Gobi Ree  On December 15, 2011 at 2:01 am

    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 |G/C_G(H)| divides 203.

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