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Suppose that $f$ is a group homomorphism from $\mathbb Z_7\times\mathbb Z_7$ to itself satisfying $f^5 = \operatorname{id}$ (where $f^5=f\circ f\circ f\circ f\circ f$). Show that $f$ is the identity.

Yeyeye
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2 Answers2

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One can think of $f$ as a $2\times 2$ matrix $A$ with entries in $\mathbb Z_7$. Since $f^5$ is identity one gets $A^5=I_2$. This shows that the minimal polynomial of $A$, denoted by $m_A$, divides $X^5-1$. But $$X^5-1=(X-1)(X^4+X^3+X^2+X+1)$$ and $X^4+X^3+X^2+X+1$ is irreducible over $\mathbb Z_7$ (why?). Thus we have $m_A=X-1$ and therefore $A=I_2$, that is, $f$ is the identity.

  • Why can you think of $f$ as a $2\times 2$ matrix with entries in $\Bbb Z_7$? – MJD Jun 04 '13 at 14:43
  • I think $f((1,0))$ and $f((0,1))$ are the four elements in the $2\times 2$ matrix. – Yeyeye Jun 04 '13 at 14:48
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    A group homomorphism $f:\mathbb Z_7\times\mathbb Z_7\to \mathbb Z_7\times\mathbb Z_7$ is also an $\mathbb Z_7$-vector space homomorphism. –  Jun 04 '13 at 14:48
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Hint:

  • In a group $G$ (what group? Its order divides $42\cdot 48$) if $a\in G$ then $a^{|G|}=\mathrm{id}$.
  • If $a^k=\mathrm{id}$ and $a^n=\mathrm{id}$ then $a^{\gcd(k,n)}=\mathrm{id}$.

Can you apply the above to your problem? What if $f^3=\mathrm{id}$ or $f^{11}=\mathrm{id}$?

P..
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  • Is the group $G$ of all group homomorphisms? why dose its order divide $42\cdot 48$? – Yeyeye Jun 04 '13 at 14:52
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    Since the fifth power of the homomorphism is identity, it’s an automorphism, and thus is contained in the automorphism group of the dimension-two vector space in question. This automorphism group is $\mathrm{GL}_2(\mathbb F_7)$. A simple counting argument shows that this group has order $(7^2-1)(7^2-7)$. (I must say that I find this answer very slick and very satisfying!) – Lubin Jun 04 '13 at 15:08