Show that $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ has seven subgroups of order $2$.
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6Count the elements of order 2. Each is the generator of a subgroup. And each element of your group is of order 2, except the identity. – Jean-Claude Arbaut Dec 07 '13 at 00:34
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$([0]; [0]; [0])$; $([1]; [0]; [0])$; $([0]; [1]; [0])$; $([1]; [1]; [0])$; $([0]; [0]; [1])$; $([1]; [0]; [1])$; $([0]; [1]; [1])$; $([1]; [1]; [1])$:
Now $[0]+[0] = [1]+[1] = [0]$ in $\mathbb Z_2$, and the direct sum construction is defined with operation: $$(x_1; y_1)(x_2; y_2) = (x_1x_2; y_1y_2):$$ Thus, if we take any $a = (x; y; z)$ element in $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ then we must have $a_2 = (x; y; z)(x; y; z) = (x + x; y + y; z + z) = ([0]; [0]; [0])$, the identity of $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$.
There are seven elements of $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$ of order 2 (every element except $e$), and for each such $a$ there is a subgroup of order 2, namely $fe; ag$. This gives seven different subgroups.
However, this is all of the subgroups of order 2, since a subgroup of order 2 has $e$ and one other element.
