$F_2×F_2$的展示

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UTM groups and symmetry, Armstrong

27.7. Show that $F_2 \times F_2$ is not a free group. Write down a presentation for $F_2 \times F_2$.

使用直積的展示「若$G=〈S∣R〉,H=〈T∣Q〉$，且$S ∩ T = \{e\}$，則$G × H=〈S,T∣R,Q, [S,T]〉$」解决如下
首先将两个 $F_2$ 嵌入到一个群 $F_4=〈a, b, c, d〉$ 中$〈a, b〉∩〈c, d〉= \{e\}.$
则$F_2 × F_2 = 〈a, b, c, d \mid [a,c],[a,d],[b,c],[b,d]〉,$右边是4个交换子

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Example 2.4说1^#的展示有错！

Note that if $GX_1$ and $GX_2$ are free groups, the already known result for subgroups of direct products of two free groups is achieved.
Stallings showed that $F_2 × F_2 = 〈a, b, c, d \mid [a,c],[a,d],[b,c],[b,d]〉$ is not coherent by checking that the kernel of $F_2 × F_2 → ℤ$ sending $a, b, c$ and $d$ to 1 is finitely generated but not finitely presented.
Let $K$ be that kernel. Proving that $K$ is not finitely presented is a consequence of the previous theorem: It is easy to check that $L_1$ and $L_2$ are $〈〈ab^{−1}〉〉$ and $〈〈cd^{−1}〉〉$, respectively. These are not finitely generated, so by Theorem 2.2, the result is obtained.

$L_1$和$L_2$是什么? 唉还得用到上面的Main theorem

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为什么1^#的展示有错

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"is not coherent"是有错的意思吗