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Basic Subgroup Conditions

could someone please explain how the one step subgroup test works,

I know its important and everything but I do not know how to apply it as well as with the two step subgroup.

If someone could also give some examples with it it would be really helpful.

thank you


marked as duplicate by Gerry Myerson, Brandon Carter, t.b., user31373, Zhen Lin Aug 16 '12 at 10:20

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Rather than prove that the "one step subgroup test" and the "two step subgroup test" are equivalent (which the links in the comments do very well), I thought I would "show it in action".

Suppose we want to show that $2\Bbb Z = \{k \in \Bbb Z: k = 2m, \text{for some }m \in \Bbb Z\}$ is a subgroup of $\Bbb Z$ under addition.

A) The "two-step method": first, we show closure - given $k,k' \in 2\Bbb Z$, we have that:

$k = 2m,k' = 2m'$ for some integers $m,m'$, so $k+k' = 2m+2m' = 2(m+m')$. Since $\Bbb Z$ is a group, and closed under addition, $m+m'$ is an integer, so $k+k' \in 2\Bbb Z$.

Next, we show that if $k \in 2\Bbb Z$, $-k \in 2\Bbb Z$: since $k = 2m$, for some integer $m$, we have $-k = -(2m) = 2(-m)$, and since $-m$ is also an integer, $-k \in 2\Bbb Z$.

B) The "one step method": here, we combine both steps into one: given $k,k' \in 2\Bbb Z$, we aim to show that $k + (-k') \in 2\Bbb Z$. As before, we write:

$k + (-k') = k - k' = 2m - 2m' = 2(m -m')$, and since $m - m'$ is an integer, $k + (-k') \in 2\Bbb Z$.

A more sophisticated use of this test, is to show that for any subgroup $H$ of a group $G$, and any element $g \in G$, $gHg^{-1} = \{ghg^{-1}: h \in H\}$ is also a subgroup of $G$. So given any pair of elements $x,y \in gHg^{-1}$, we must show $xy^{-1} \in gHg^{-1}$. Note we can write:

$x = ghg^{-1}$, for some $h \in H$, $y = gh'g^{-1}$, for some $h'\in H$.

Then $y^{-1} = (gh'g^{-1})^{-1} = (g^{-1})^{-1}h'^{-1}g^{-1} = gh'^{-1}g^{-1}$, so:

$xy^{-1} = (ghg^{-1})(gh'^{-1}g^{-1}) = gh(g^{-1}g)h'^{-1}g^{-1} = gh(e)h'^{-1}g^{-1} = g(hh'^{-1})g^{-1}$.

Since $H$ is a subgroup, it contains all inverses, so $h'^{-1}$ is certainly in $H$, and $H$ is also closed under multiplication, so $hh'^{-1} \in H$, thus:

$xy^{-1} = g(hh'^{-1})g^{-1} \in gHg^{-1}$, and we are done.

  • 4
    $\begingroup$ What bothers me about the "one-step" method is that surely one also has to prove that the purported subgroup is non-empty, so it's really 2 steps (and the "two-step" method is really 3 steps). Of course, it's usually obvious that the set isn't empty, as in your example, $2{\bf Z}$, but it still ought to be noted. $\endgroup$ – Gerry Myerson Aug 15 '12 at 4:38
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    $\begingroup$ you guys are awesome that really helped me out thanks again $\endgroup$ – user37012 Aug 15 '12 at 17:04

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