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$H$ is a subgroup of $G$ so that $G:H=5$

$ a \in Z(G) , ord(a)=3$

Show that $ a \in H $

What I did:

if $ a \notin H $ then:

$ G/H = \{ H, aH, a^2H, gH, agH \} $ for some $ g \in G-H $

But I don't know how to procceed...

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Try showing that if $gH$ is different from $H, aH, a^2H$, then all 6 of $a^ig^{\epsilon}H$ are all distinct, with $0\leq i<3, 0\leq \epsilon <1$ – Aaron Jun 17 '12 at 6:37
Well, this is what I've "proved": $ H, aH and a^2H $ are clearly distinct. Because $ G:H = 5 $ then there must be $ g \in G-H $ so that $ gH $ is distinct. now we prove that $ agH $ is distinct: if $ agH = H $ then $ gH = a^2H $ - not possible if $ agH = aH $ then $ gH = H $ - not possible if $ agH = a^2H $ then $ gH = aH $ - not possible now we prove that $ agH $ is distinct: if $ a^2gH = H $ then $ gH = aH $ - not possible if $ a^2gH = aH $ then $ gH = a^2H $ - not possible if $ a^2gH = a^2H $ then $ gH = H $ - not possible if $ a^2gH = agH $ then $ agH = gH $ - not possible – catch22 Jun 17 '12 at 8:34
But in this proof, I didn't use the fact that $ a \in Z(G) $! – catch22 Jun 17 '12 at 8:39
up vote 2 down vote accepted

As in Cihan's answer, let $A = \langle a\rangle$. Since $a \in Z(G)$, $AH$ is a subgroup of $G$ containing $H$. As $[G : H]$ is prime, $AH = H$ or $AH = G$. In the latter case, $G = H\cup aH \cup a^2H$, a union of at most 3 left cosets of $H$, contradicting $[G:H] = 5$. So $AH = H$, which implies $a \in H$.

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Can you please explain why : As $[G:H]$ is prime, $AH=H$ or $AH=G$ – catch22 Jun 17 '12 at 9:46
+1: Much neater :) – Cihan Jun 17 '12 at 11:31
In general if $a \in N_G(H)$ (the normalizer of $H$ in $G$) with $gcd(ord(a), [G:H])=1$, then $a \in H$. – Nicky Hekster Jun 17 '12 at 15:02
@Serkan, in your example $a$ as order 3 and $H$ has index 3. So it is not a counterexample. – Nicky Hekster Jun 18 '12 at 8:39
Let me give you a proof. Look at $K=H\langle a \rangle$. Since $a$ normalizes $H$, $K$ is a subgroup and $H$ is normal in $K$. Hence $index[K:H] = index [\langle a \rangle: H \cap \langle a \rangle]$. Since $index[K:H]$ divides $index[G:H]$ and $gcd(order(a), index[G:H])=1$, it follows that $K=H$, whence $a \in H$. – Nicky Hekster Jun 18 '12 at 8:50

Suppose that $a\notin H$, and let $H,aH,a^2H,g_1H$, and $g_2H$ be the left cosets of $H$. Clearly left multiplication by $a$ permutes $H,aH$, and $a^2H$. It also permutes the other two cosets: $ag_1H$ must be $g_1H$ or $g_2H$, and $ag_1H=g_1aH\ne g_1H$, since $a\in Z(G)$ and $aH\ne H$, so $ag_1H=g_2H$. Similarly, $ag_2H=g_1H$. But then $g_1H=a^3g_1H=a^2g_2H=ag_1H=g_2H$, which is a contradiction.

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@Serkan: Shows I need to read all comments before posting... that's pretty much what I just wrote. – Arturo Magidin Jun 17 '12 at 19:22

$G$ acts on the left cosets of $H$ by left multiplication; this gives a map from $G$ to $S_5$. The action of $a$ has exponent $3$, so it is either trivial or else corresponds to a $3$-cycle. In particular, it must fix some coset, $gH$, so that $agH = gH$. This implies that $g^{-1}agH = H$, but since $a$ is central, $g^{-1}ag = a$. Therefore, $aH=H$, so $a\in H$.

This generalizes to non-prime index cases:

Proposition. Suppose $H$ is a subgroup of $G$ of index $n$, and let $p$ be a prime such that $\gcd(n,p)=1$. If $a\in Z(G)$ has order $p$, then $a\in H$.

Proof. Let $G$ act on the left cosets of $H$ by left multiplication; this induces a homomorphism $G\to S_n$, and the image of $a$ has exponent $p$. Writing the image of $a$ as a product of disjoint cycles, we see that each cycle has length either $1$ or $p$. Since $p$ does not divide $n$, the image of $a$ must have a fixed point. Therefore, there exists a coset $gH$ such that $agH = gH$, which implies $g^{-1}agH=H$. Since $a$ is central, $g^{-1}ag=a$, so we have $aH=H$, hence $a\in H$, as claimed. $\Box$

Added. As Serkan notes in a comment I only saw after posting the above, the hypotheses of the proposition can be further weakend to let the order of $a$ be $m$, where every expression of $n$ as a sum of divisors of $m$ must involve at least one summand equal to $1$.

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Let $Z = Z(G)$. Since $ZH = HZ$, $ZH$ is a subgroup of $G$. We have $$5 = |G:H| = |G:ZH||ZH:H|$$ so either $|G:ZH| = 1$ or $|ZH:H| = 1$. In the latter case, we get $ZH = H$; hence $Z \subseteq H$ and we are done.

In the former case, $G = ZH$. So $$|Z : Z \cap H| = |ZH:H| = |G:H| = 5$$ Let $A = \langle a \rangle$. We know $A \subseteq Z$. Since $Z$ is an abelian group, $A(Z \cap H)$ is a subgroup of $Z$. Now we have $$5 = |Z : Z \cap H| = |Z : A(Z \cap H)||A(Z \cap H) : Z \cap H|$$ If $A \subseteq Z \cap H$, we are done. So suppose $A(Z \cap H) > Z \cap H$. Then by above, we get $Z = A(Z \cap H)$ and $|Z : Z \cap H| = 5$. But then $$|A: A \cap Z \cap H| = |A(Z \cap H): Z \cap H| = 5$$ and this contradicts $|A| = 3$.

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