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Math Help - Algebra, Problems For Fun! (4)

  1. #1
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    Algebra, Problems For Fun! (4)

    Ok, for this one you only need some very basic knowledge of group theory:

    Let G be a finite non-abelian group and |G|=n. Let k be the number of conjugacy classes of G. Prove that k \leq \frac{5n}{8} with equality if and only if |Z(G)|=\frac{n}{4}.
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  2. #2
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    Quote Originally Posted by NonCommAlg View Post
    Ok, for this one you only need some very basic knowledge of group theory:

    Let G be a finite non-abelian group and |G|=n. Let k be the number of conjugacy classes of G. Prove that k \leq \frac{5n}{8} with equality if and only if |Z(G)|=\frac{n}{4}.
    Is Z(G) the center of G? (I should know this, but it has been awhile.)
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  3. #3
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    Quote Originally Posted by chabmgph View Post

    s Z(G) the center of G?
    yes.
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  4. #4
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    Quote Originally Posted by NonCommAlg View Post
    Ok, for this one you only need some very basic knowledge of group theory:

    Let G be a finite non-abelian group and |G|=n. Let k be the number of conjugacy classes of G. Prove that k \leq \frac{5n}{8} with equality if and only if |Z(G)|=\frac{n}{4}.
    I will use the fact that the smallest non-abelian group has order 6, which is not hard to show.

    Let  C(g_1),C(g_2),...,C(g_k) be the conjugacy classes of G. Assume WLOG that the first m are non-central.

    Suppose |Z(G)|\not\le{\frac{n}{4}}. Then  |G/Z(G)|\lneq{4} Thus  |G/Z(G)|= 3,2, or 1. We will reach a contradiction for each case.

    The case  |G/Z(G)|=1 is impossible since this would imply G were abelian. So assume  |G/Z(G)|=2 .

    Let g_i\in{G} with  i\le{m}. Let  g\in{G\setminus{Z(G)}}  with g\neq{g_i}. Consider  gg_{i}g^{-1}

    Since g and g_i are both noncentral and G/Z(G) has order two we know that gg_i\in{Z(G)}

    Thus gg_ig^{-1} = g^{-1}gg_i=g_i
    Hence, we have shown that  gg_ig^{-1} = g_i for all  g\in{G} But this implies  g_i\in{Z(G)} which is a contradiction.

    I don't have time to complete the last case. But it is fairly similar. Again use the fact that we know the group structure of  G/Z(G) and find the similar contradiction. Then use the class equation to get the inequality for k. I have not proven the case of equality but I'll leave that for someone else.
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  5. #5
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    Quote Originally Posted by bulls6x View Post
    I will use the fact that the smallest non-abelian group has order 6, which is not hard to show.

    Let  C(g_1),C(g_2),...,C(g_k) be the conjugacy classes of G. Assume WLOG that the first m are non-central.

    Suppose |Z(G)|\not\le{\frac{n}{4}}. Then  |G/Z(G)|\lneq{4} Thus  |G/Z(G)|= 3,2, or 1. We will reach a contradiction for each case.

    The case  |G/Z(G)|=1 is impossible since this would imply G were abelian. So assume  |G/Z(G)|=2 .

    Let g_i\in{G} with  i\le{m}. Let  g\in{G\setminus{Z(G)}} with g\neq{g_i}. Consider  gg_{i}g^{-1}

    Since g and g_i are both noncentral and G/Z(G) has order two we know that gg_i\in{Z(G)}

    Thus gg_ig^{-1} = g^{-1}gg_i=g_i
    Hence, we have shown that  gg_ig^{-1} = g_i for all  g\in{G} But this implies  g_i\in{Z(G)} which is a contradiction.

    I don't have time to complete the last case. But it is fairly similar. Again use the fact that we know the group structure of  G/Z(G) and find the similar contradiction. Then use the class equation to get the inequality for k. I have not proven the case of equality but I'll leave that for someone else.
    you almost got the idea but you need to put everything together. a quick way to show that |G/Z(G)| \geq 4 is to recall that if G is non-abelian, then G/Z(G) is never cyclic. now you need to

    combine this result with the class equation to prove the first part. the proof of the second part of the problem comes from the proof of the first part.
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